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THE 


KSSENTIALS OF ALGEBRA 
FOR SHCONDARY SCHOOLS 


BY 


ROBERT J.. ALEY, Pa.D. ra 


PROFESSOR OF MATHEMATICS, INDIANA UNIVERSITY "¥ 


AND @ 


DAVID A. ROTHROCK, Pu: 


ASSOCIATE PROFESSOR OF MATHEMATICS, INDIANA YNIVERSITY 





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fa Xx w7 C6 uATHEMATICS LIBRARY 


PREFACE. 


In the preparation of this book the authors have made 
an earnest effort to retain all the essentials of the older 
Algebra text-books, and to introduce and properly empha- 
size certain newer features which the mathematical studies 
of the present demand. 

The following are some of the special characteristics of 
the book: 

1. The Number System. ‘The number system is pre- 
sented in the first chapter, and from the arithmetical 
system extension is made to the algebraic number. system. 
—. In this way the idea of negative number is introduced 
— and the fundamental operations are explained. 
~~ 2. Factoring. This subject is treated with particular 
3 fullness, and use is made of the factorial method wherever 
_s-applicable in the study of Algebra. At the first reading, 

Sections 79, 80, and 81, covering certain details of factor- 

ing, may be omitted if thought desirable. The ordinary 

student, however, should have no special difficulty in 
mastering these sections. 

3. The Graph. The work with graphs is made an in- 
tegral part of the book. The graphs of simple and 
quadratic equations are used freely to aid the pupil’s 
understanding of the solutions involved. Graphic illus- 
trations are given wherever it is thought they will make 


the subject clearer. 
iii 


334.356 


lv PREFACE. 


4. Type Forms. Type forms play an important part in 
the study of Algebra. ‘The work of the student is greatly 
simplified if he learns early in his course to recognize and 
to understand these types. ‘Type forms are extensively 
used in multiplication, division, factoring, and equations. 

5. Exercises. ‘The exercises have been selected with a 
view of clarifying the text and enforcing fundamental 
_principles. ‘They are numerous, and are difficult enough 
to call for effort on the part of the student. 

It is believed that the book contains sufficient matter 
to furnish a thorough training in the elements of Algebra 
and to meet the entrance requirements of American 
colleges. 


CHAPTER 


TH 
{Fi 
i8ae 


XVII. 
XVIII. 
348.4 


CONTENTS. 


INTRODUCTION F ; ‘ : 5 : ‘ ; 
DEFINITIONS . : : , ; . : ; : 
ADDITION AND SUBTRACTION. ; : A : 
MULTIPLICATION AND DIVISION . ‘ : é : 
IMPORTANT IDENTITIES : ; , : A : 
FACTORING . ‘ ; : : : ri a : 
Divisors AND MULTIPLES . : ; ; ; ; 
FRACTIONS . ; : : , ; : : ; 
EQUATIONS IN ONE VARIABLE . - 

LINEAR EQUATIONS IN Two VARIABLES . ‘ : 
SIMULTANEOUS EQUATIONS . 

EVOLUTION . 2 Cie : 

THEORY OF INDICES 

Rapicats, SurRDs, AND IMAGINARIES 

QUADRATIC EQUATIONS IN A SINGLE VARIABLE 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS 
RATIO, VARIATION, AND PROPORTION : ‘ 
PERMUTATIONS AND COMBINATIONS . ; ‘ P 
SERIES . : : ; : ‘ : : ‘ : 
LOGARITHMS . : : - ‘ ‘ i : : 


PAGE 


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THE 
ESSENTIALS OF ALGEBRA. 


CHAPTER I. 
INTRODUCTION. 


1. The Integral Number System is that orderly succession 
by ones which we first learn by counting. We are familiar 
with it in the Arabic numeral form of 1, 2, 3, 4, 5, 6, T, 8, 
Pave and so on. The characters 1, 2, 3, 4, etc., are 
symbols of number, but we shall hereafter, by the use of a 
common figure of speech, speak of them and other number 
symbols as number. 


2. Elementary Number Notions. We know that 3+4=7, 
because by counting 3 and then 4 more we reach 7. This 
may be seen by counting these groups, 

@eo 6880806 @ 


All the results of addition are primarily determined by 
counting. In practice, a number of simple addition results 
are determined by counting, and then these are made a 
matter of memory. 3 x 4= 12, because by counting 3 
groups of 4 each we reach 12. This is seen in the follow- 
ing arrangement : 


The truth of a multiplication table is also established by 
counting. The number system shows that 34+4=4+3; 
1 


2 THE ESSENTIALS OF ALGEBRA. 
for by counting 3 and then 4 more, we reach the same 
result as by counting 4 and then 3 more. 


3 x4=4 x3 because 3 groups of 4 each meas the same 
sum as 4 groups of 3 each. 


Ie 
Be uly 3 


Illustrations enough have been given to show that the 
integral number system is the real basis of the funda- 
mental parts of arithmetic. 


3. Fractions in the Number System. As long as no exact 
measurements are needed, nor accurate divisions attempted, 
the integral number system is sufficient. If a stick is 
more than 9 inches and less than 10 inches in length, we 
can not express its exact length by means of the integral 
number system. A similar difficulty arises in attempting 
to answer the question 8+ 3 = what? To answer all such 
questions, fractions have been devised and made a part 
of the number system. The addition of fractions to the 
number system made possible many arithmetical opera- 
tions which were before impossible. The field of arith- 
metic was thus greatly enlarged. 


4. Incommensurables in the Number System. When the 
necessity for extracting roots arose in the development of 
arithmetic, it was found that many roots could not be ex- 
actly determined. For example, the square root of 2 lies 
between 1 and 2, between 1.4 and 1.5, between 1.41 and 
1.42, between 1.414 and 1.415, etc. We may extend this 


INTRODUCTION. 3 


process of locating the square root of 2 between consecu- 
tive numbers of the number system as far as we please, 
but we can never find its exact value. Such numbers as 
the square root of 2, and the square and cube roots of other 
numbers which can not be exactly found, are called incom- 
mensurable numbers or merely tncommensurables. Although 
such numbers can not be exactly expressed, the number 
system now includes them. 


5. Numerical Arithmetic Complete. With the number 
system so developed as to include integers, fractions, and 
incommensurables, ordinary numerical arithmetic is com- 
plete. This means that in performing the operations of 
ordinary arithmetic no necessity arises for any other kind 
of numbers. 


6. Literal Arithmetic. In percentage we frequently 
represent the base by 6, the rate per cent by 7, the per- 
centage by p, the amount by a, and the difference by d. 
When we do this, we can transform the rules for the cases 
of percentage into the following forms : 


Cen Ut. 
(2) r=p+b. 
(3) b=p+r. 


(4) a=b+bx?r. 
(ey ia D9, 


The symbols 08, 7, p, a, and d may be considered as 
particular numbers of the number system. When thought 
of in this way, they are mere abbreviations of numbers. 
Since they may be the abbreviations of any numbers 
whatsoever, we may think of the symbols themselves as 


4 THE ESSENTIALS OF ALGEBRA. 


numbers. When a symbol, such as any of the above, is 
thought of in this way, it is called a general number. Such 
a number is frequently called a literal number. These 
symbols of general or literal numbers may have particular 
numerical values assigned to them. In order to find 8% 
of 250 we take form (1), on page 3, and put 250 instead 
of 6, and .08 instead of 7. We then have 


p= OX T= 200 vo —aue 


7. Substitution. Zhe process of putting a particular num- 
ber in the place of a general one ts called substitution. 

By substitution all the results of general or literal 
arithmetic become particular. The solution of a problem 
in ordinary arithmetic is a mere matter of substituting 
particular numbers for general ones in the proper literal 
form, as is illustrated in the percentage problem of 
Section 6. 

The area of a rectangle is the product of its length and 
width. If we represent area by a, length by J, and width 
by w, we at once have the general form 


a=Ilxw. 


If we wish to find the area of a lot 66 feet long and 
30 feet wide, we put 66 for 7 and 30 for w, and we have 


a=lxw=66 x 30= 1980. 


8. Algebraic Expression. Any combination of literal num- 
bers or of literal and arithmetical numbers by means of any 
or all of the signs of addition, subtraction, multiplication, 
division, involution, and evolution 1s an algebraic expression. 

a +y-—z is an algebraic expression and is read a plus 
y minus z The algebraic expression a x b—e+d+4is 

/ 


INTRODUCTION, § 


read a times 6 minus e divided by @ plus 4. The word 
function is frequently used instead of expression. The 
parts of an algebraic expression separated by either of 
the signs + or — are called terms. In ax + by — cyz, there 
are three terms; viz., ax, by, and cyz. 


9. Signs used in Algebraic Expressions. ‘The signs +, 
—, X, +, and +/ are used as in arithmetic. They denote 
addition, subtraction, multiplication, division, and root 
extraction, respectively. Multiplication is also indicated 
by a dot (-), and by writing the characters adjacent to 
each other. ax 6, a:b, and ab mean exactly the same 
thing; viz., @ multiplied by 6. Between arithmetical 
numbers also the signs x and - are used to denote multi- 
plication. The multiplication of an arithmetical and literal 
number or of two literal numbers is denoted by writing 
them consecutively. 56 means 5x6. ab means a x 6. 
6ab=6xaxb=6-a-b. The first form (646) is the 
one generally used and is read szz ab. 

There are five forms of division sign in general use, 
6+ 2, 2)6, 6:2, §, and 6/2, all meaning exactly the 
same thing; viz., 6 divided by 2. In algebra the 
fourth and fifth forms are more frequently used than 
the others. 

Vz, Va, Vb are read square root of a, cube root of a, 
and fourth root of 6, respectively. 

a*, 68, z* are read a square, 6 cube, and x to the fourth 
power, respectively. 


a=aa; 6=bbb; 2t= rrr. 


In the above expressions the 2, 3, and 4 are called 
exponents. 


So 


THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 
Read the following algebraic expressions: 
1 a+b—c+4. 2. 4a—3b+<ab. 
3. ab —4ac+3.ad —*. 
; is read b divided by c, or 6 over ec. 


The last term of (3) should be read minus the quotient b divided 
by c, or minus the fraction b over c. ° 


4. 2a+b+5abe —A4 be. 6. 5a®—A aa? +Vab. 
= 3 Baie: 
5. a —4ar+vo0. 7. 6 abt — + Va 


8. 5a ern, 
y 


/ ee . 
5a vu is read 5a times the square root of z. 


9. nye +87 17 vay 10. Sat Vyz 24 8a Von, 


* 10. Precedence of Signs in Algebraic Expressions. Jf only 
the signs + and — occur in an algebraic expression, the 
operations are to be performed in order from left to right. 


For example: 8—4+4+3+42—5=4. 


If only the signs x and + occur in an algebraic expres- 
sion, the operations are to be performed in order from left 
to right. 


For example : 6+38x4+2=4. 


If the signs +, —, X, and + occur in an aigebraie ex- 
pression, the multiplications and divisions are first per- 
formed, and then the additions and subtractions. 


INTRODUCTION. T 
For example: 44+ 3x 2—12+6x342x4. 


Performing the multiplications and divisions in the above, 
we have 4+6—6+48. Now performing the additions and 
subtraction, we get 12, which is the value of the expression. 


* 11. Signs of Aggregation. ‘The signs of aggregation or 
grouping are the parentheses ( ), braces { }, brackets [ ], 
and vinculum or bar~——. Each of these signs indicates 
that the expression within or under it must be treated as 
a whole. (6c — ad)+ 6 means that the difference between 
be and ad is to be divided by 6. The expressions (be — 
ad)+6b, {be—adj+b, [be—ad]+6, and be—ad+6 all 
mean precisely the same thing. ‘The four signs of aggre- 
gation are all called by the general name parentheses. ‘The 
different forms are necessary to avoid confusion when one 
or more groups are included within another group. 





For example: 5 x {12+(7 x64+2+4+[2x3+41])}. 


This becomes 5 x $12 +(7 x 8+4+7)}, which in turn 
becomes 5 x §12.+ 23, or 5x6=80. 

Multiplication of a quantity within a parenthesis by any 
quantity is indicated by writing the multiplier before 
or after the parenthesis. 5(a+6) means 5x(a+6). 
(a+6)5 means (a+ 6) x d. 


* 12. Coefficient. In the expressions 5a, 32, and Ty, 5, 3 
and 7 are the coefficients of a, x, and y, respectively. 

In an indicated product any factor or factors may be 
considered the named part, then all the other factors con- 
stitute the coefficient. Thus, in 8 aay, 8 is the coefficient 
of axy, 8 a is the coefficient of xy, and 8 az is the coefficient 


8 THE ESSENTIALS OF ALGEBRA. 


of y. In the first case ary is the named part, in the second 
xy, and the third y. 

When no numerical coefficient precedes a literal expres- 
sion, the coefficient 1 is understood. 


Read the coefficients of y? in the following expressions: 


$a, Says Lia, Ail Ua 


% EXERCISES IN SUBSTITUTION. 


Find the value of each of the following literal expressions, 
in which a=4, P=2,¢=— 3, d=), 0—=6,y— yee 


1. a+b—c+ ay. 


Substituting the values given to the above letters, this expression 


becomes 


‘In a similar manner determine the values of the following 
expressions: 


2 a+e2—yt+3d. oD Vatpete 

3. ax+tbyte. 12. V6a+vVy. 

4. «+y+a. 13. 5a+6b—4e. 

5. ax? + by? +d. 14. 38a—4b406 

ROUNE 15. 32—4y+10. 
bd 16. («+ y)2+ (y+z2)a. 

7. e+y+2+d. 17. abe + xyz. 

8. x —y +4. x 18. ax —by+cz—8 abe. 

9. w+22-+1. 19. + y+22—38 wyz. 


10. a? +2 ay + 7%. 20. y +cu + (d—Va) +46. 


INTRODUCTION. 9 


13; Use of General Number in Arithmetical Problems. All * 
the problems of ordinary arithmetic may be made general 
by the use of general or literal numbers in place of the 
arithmetical numbers involved. 


1. If John has 10 cents and Henry 12 cents, they together 
have (10 + 12) cents. This becomes general by stating it thus: 
if John has a cents and Henry 0 cents, they together have 
(a + bd) cents. 


2. If Mary is 10 years old and Susie is w years older, then 
Susie is (10 +2) years old. 


3. Ifa merchant sells a bushels of corn at b cents a bushel, 
and c bushels of wheat at d cents a bushel, and divides the 
money received equally among his e children, each one will 
receive | (ab + cd) + e] cents. 

4. A man has a cents and } dimes. How many cents 
has he? 7 

5. A man is a years old. His son is 4 as old. What is 
the combined age of father and son ? 


6. I traveled b miles at c cents a mile, and d miles at e 
cents a mile. How far did I travel and what did it cost 
me ? 

7. A rectangle is 2 rods long and y rods wide. How many 
acres does it contain ? 


8. A farm a rods long and 0 rods wide is sold at ¢ dollars 
per acre. Find the amount for which the farm sold. 


9. How many hours will be required to travel # miles if 
one third the distance be traveled at a miles per hour and the 
remaining two thirds at } miles per hour ? 


10. A man has # dollars in cash; 6 men owe him each y 
dollars, c men owe him each z dollars. How much money 
would he have if his collections were made ? 


10 THE ESSENTIALS OF ALGEBRA. 


11. A rectangle is a rods long and b rods wide. Find the 
length of its diagonal. 


12. How many hours will be required for a train running 30 
miles per hour to travel @ miles, if it makes n stops of 6 min- 
utes each ? 


13. Find the cost of excavating a basement a feet long, 
b feet wide, and ¢ feet deep at a cents per cubic yard. 


14. A man sold from a flock of n sheep; the mth part of 
them for p dollars each. He then increased his flock by a sheep 
and sold the whole lot at « dollars per head. How much did 
_ he receive for the whole flock ? 


14. Opposite Numbers. On the scale of a thermometer, 
temperature is marked both ways from 0. On the centi- 
gerade thermometer, temperatures above freezing 
read from 0 up, and temperatures below freezing 
read from 0 down. Longitude is measured both 
east and west from a fixed 0 or prime meridian. 
Latitude is measured both north and south from 
the equator. We may consider direction along 
a line to the right or to the left. Rotation may 
be opposite to that of the hands of a clock, or 
it may be clockwise. Numbers which in some 
way indicate such opposites are called Opposite 
Numbers. The need of opposite numbers becomes 
apparent when we try to generalize the opera- 
tions of arithmetic. a@—6 indicates the sub- 
traction of 6 from a. 

If a=10 and d= a a—b=10— 9=1, which 
shows:that 9 is 1 less than 10. 

If a=10 and 6=10, a—b= 10 —10 = 0, which shows 
that 10 is 0 less than 10, or that 10 is equal to 10. 


Centigrade Thermometer 





INTRODUCTION, Tel 


If a=10 and 6=11, a—b=10—11, which, what- 
ever it may be, ought to show that 11 is 1 greater than 
10. 

If a man has $500 and is in debt $400, he is worth 
$ 500 — $400 = $100. 

If a man has $500 and is in debt $500, he is worth 
$500 — $500 = 0. | 

If a man has $500 and is in debt $600, he is worth 
$500 —$600. This statement harmonizes with the state- 
ments made in the other two cases. What does it mean? 
We may interpret it by saying that he owes $100 more 
than he is worth, or that his liabilities exceed his assets by 
$100, or that he is worth $100 less than nothing. 


15. Negative Number. Such questions as the above are 
answered by the extension of the number system so as to 
include negative number. We may think of the arithmeti- 
cal number system as starting at 0 and extending indefi- 
nitely in a horizontal line to the right. It is an easy 
matter to think of a similar system extending indefinitely 
to the left from 0. These appear as follows: 


200 .--100---60.--50--4382101234..-50---60---100---200 
(ES EN ma a a SL a 

This extension doubles the scope of the number system. 
Integers, fractions, and incommensurables are all included 
in the extension to the left. 


16. Positive and Negative. That part of the number 
system to the right of 0 is called positive. The positive 
character of a number is indicated by the use of a + sign 
before it. +4, +a, and +2? are positive numbers. In 


12 THE ESSENTIALS OF ALGEBRA. 


practice the + sign is frequently omitted, so that the ab- 
sence of a sign before a number shows that it is positive. 

That part of the number system to the left of 0 is called 
negative. ‘The negative character of a number is indicated 
by the use of a — sign before it. — 6a, —a?,and — 112*y 
are negative numbers. 

The important idea in the two parts of the number sys- 
tem is that of oppositeness. When anything is represented 
by a positive number, its opposite is represented by a 
negative number. If time A.D. is positive, then time B.C. 
is negative. If distance to the right is positive, then dis- 
tance to the left is negative. 


ILLUSTRATIVE EXERCISES. 


1. If two points are on the same meridian in latitude + 30° 
and — 20°, respectively, how far apart are they? This means 
that one is in north latitude 30°, and the other in south latitude 
20°. They are evidently 30° + 20° = 50° apart. 

Draw a diagram illustrating this. 

2. On a certain day the lowest temperature recorded was 
— 5° and the highest +12°. What was the difference in tem- 
perature between the lowest and highest ? 

3. A man was born in the year —31 and died in the year 
+43. How old was he ? 

4. A man travels +45 miles from A, and his friend travels 
— 80 miles from A. How far apart are they ? 

5. Two places are in —63° and +87° longitude, respectively. 
How far apart are they ? 


17. Extension of Meaning of Negative. In the above 
exercises and illustrations we have thought of negative 
number as beginning at 0 and extending in the opposite 
direction from that of positive number, which also begins 


INTRODUCTION. 13 


at 0. Beginning at zero is not a necessary part of the 
meaning. The necessary part is that of oppositeness. 
Number starting anywhere is negative if it denotes exten- 
sion in the opposite direction to that of positive number. 


A man travels east 12 miles, then west 4 miles, then east 11 miles, 
then west 15 miles. If we select east as the positive direction, then 
west is the negative direction. We may then say that a man travels 
+12 miles, then — 4 miles, then +11 miles, and then — 15 miles; 
as shown in the following diagram. 


+ 12 mi. 





18. Double Use of the Signs + and —. The signs + and 
— in algebra retain their arithmetical sense, and denote 
addition and subtraction, respectively. When used in 
this sense they are called signs of operation. The signs 
+ and — are also used to denote the quality of a number, 
that is, to indicate that the number belongs to the posi- 
tive or negative part of the number system. In this 
sense, the signs + and — denote oppositeness, and are 
called segns of quality. Smaller signs slightly elevated are 
sometimes placed before a number to denote its quality. 


Pemeraiiple: 75, 7D, ta, a, 7156, 718 ¢. 


19. Algebraic Number. When the quality of a number 
is considered, the number becomes algebraic. The signs of 
algebraic number are + and —. The absolute value of a 
number is its value without regard to quality. +a, +4, 
+11, —T, are algebraic numbers. a, 6, 11, T, are their 
absolute values. 


CHAPTER il. 
DEFINITIONS. 


20. Identity. In the equality 54—2xr=5-2 we have 
merely the statement that 5 2 diminished by 2 2 becomes 3 z. 
No question need be asked concerning the number or value 
represented by z ‘The equality is true, whatever value x 
may have. If x=1, the equality becomes 5x1—2x1 
=$§xl. If27=5,it becomes5x5—2x5=3x5. te 
be any number a, it becomes 6a —2a= 3a. 


An equality, true for all values of the letters considered, 
as called an identity. 


In an identity both sides of the equality may be reduced 
to the same form. 22+ 2ar+ y2—2ar=27+ y7 is an 
identity because the left side becomes 27 + y? by uniting 
2axand — 2 az. 


21. Equation. ‘The equality x+5=7 is entirely dif- 
ferent from that considered in the preceding Section. If 
x be 1, the equality does not exist, for 1+ 3 is not equal 
to 7. Neither does it exist if x be 3, for 8+ 8 18 note 
If x be 4, the equality exists, for then we have 44+ 3=7, 
a numerical identity. We see that the equality 7+ 3=7 
restricts the value of x to 4. 


An equality which contains one or more restricted letters 
as called an equation. 
14 


DEFINITIONS. NF 


An identity is usually distinguished from an equation 
by having its sign of equality written thus, =, while the 
equation retains the sign =. The sign = is read “is 
identical with” or “identically equals.” ‘The sign = is 
read “is equal to” or “ equals.” 


5a+3a= 8a is read 5a plus 8a is identical with 8 a. 
3x%+4= 20 is read 3a plus 4 equals 20. 


22. Variables. The letters of an equality which are re- 
stricted in value are called variables. 

They are generally, although not necessarily, represented by 
the last letters of the alphabet. 

In the equality « + y= 5, w and y are variables. 


23. Constants. The letters of an equality which are not 
restricted and all arithmetical numbers are called constants. 


They are generally, although not necessarily, represented by 
the first letters of the alphabet. 
In the equality ax + by =c, the constants are a, b, and c. 


EXERCISES. 


Distinguish between equation and identity in the following 
equalities; also point out the variables and constants. 


1. 5a@—4e¢+4+a=2-2. 9. v—2ar+a74+2 ar=a7+22, 
2. 8ytv—4y+6e=4y4+7u. 10. 82—52=12. 

3. d4—5=—10. 1l. aw+ by=c. 
4°124a—304+40=9a+4b. 12. 32+427=52422. 

5. 4a+10=—18. 13. Te¢+a=12. 

6. Zav+5ar—Sar=4ax, 14. ae+5arx=a’'x(8 — 2). 
Toy fy 15, 15. 5u+9u—T x= 28. 

8. 4ay+6ary —2 vy=8 wy. 


16 THE ESSENTIALS OF ALGEBRA. 


24. The Root of an Equation. A value of the variable 
which, when substituted for the variable, reduces an equation 
to an ‘identity, ts called a root of the equation. 


The equation 2 — 5 = 8 asks what number diminished 
by 5equals 8. The answer, «= 13, is the root, for 13—5=8 
is an identity. 


The process of obtaining the value of the root is called 
solving an equation. 


25. Axioms. In solving an equation certain elementary 
facts are taken for granted; that is, their truths are ac- 
cepted without proof. Such self-evident truths are called 
axioms. ‘The three following are of use to us at present: 


(1) Numbers equal to the same number, or to equal num- 
bers, are equal to each other. 

Ex. 4+2=6,383+38=6; hence 4+2=—3+4. 

If 3a—5=10, and 2644=10, then 83a—5=26+4+4. 

(2) If equals be added to or subtracted from equals, the 
results are equal. 

Ex. 3+2=5; then8+2—2=5-—-2. 

If a+b=e,thena+6b—6b=c—6. 

If x+y=65,thnza+y+4=544. 

(3) Tf equals be multiplied or divided by equals, the re- 
sults are equal, 

Ex. 1£=3+4H. 

Multiply by 3, Ig X 8=3x3414x 3,orl0=9+4+1. 

4z 20 


If 4a = 20, then Se eee 
4 4 


DEFINITIONS. aye 


26. Solution of Exercises. The algebraic equation can be 
used to advantage in the solution of many exercises found 
in ordinary arithmetic. The process consists in first ex- 
pressing the exercise as an algebraic equation, and then 
applying the.axioms so as to find the root. This will be 
illustrated in the following exercises. 


EXERCISES. 


1. A and B have $900; A has $100 more than twice what 
B has. How much has each ? 


SOLUTION. 


Let x = B’s money. 
22 + $100 = A’s money. 
x +22 + $100 = both A’s and B’s money. 
$900 = both A’s and B’s money. 
By Axiom (1), z+22+100= 900. 
By Axiom (2), +2 z=900—100 = 800 (subtracting 100 from equals). 
oa = O00, 
By Axiom (3), 2 = 2662, B’s money. 
22+ 100 = 5334 + 100 = 6334, A’s money. 


2. A and B have $1500; A has $300 less than 3 times B’s. 
How much has each ? 


3. What number added to twice itself will make 900 ? 


4. The sum of two numbers is 84; the larger is 11 times the 
smaller. What are the numbers ? 


5. John has $300 more than Henry; they both have $2100. 
How much has each ? 


6: A house and lot cost $3700; if the house is worth $700 
more than the lot, what is the value of each ? 


18 THE ESSENTIALS OF ALGEBRA. 


7. Divide $720 among three men so that the second shall 
have twice as much as the first, and the third 3 times as much 
as the first. 

8. Divide 440 into three parts so that the second. part’ shall 
be 100 more than the first, and the third part as much as the 
sum of the first and second parts. 

9. A horse and carriage cost $288; the carriage cost 4 as 
much as the horse. How much did each cost ? 


SOLUTION. 
Let x = cost of horse. 
4a = cost of carriage. 
x++#2z=cost of both. 
$288 = cost of both. 
By Axiom (1), 2+4#x2= 288. 
By Axiom (2), 5x +42 = 288 x 45, multiplying by 5. 
9x = 1440. 
By Axiom (3), x = 1440 + 9 = $160, cost of horse. 
#x=# of 160 = $128, cost of carriage. 
10. What number increased by 2 of itself is 550 ? 
11. One third of a number increased by 4 of the number 
is 455. What is the number ? 
12. A’s money is 2 of B’s money; together they have $1300. 
How much has each ? 
13. Four times a number increased by % of the number 
is 475. What is the number ? 
14. Two numbers added together make 80; the greater is 5 
more than 4 times the lesser. What are the numbers ? 
15. If to my age you add its half and its third and 50 years 
more, the sum will be 3 times my age. What is my age? 


16. If to the double of a number you add its half and 42 
more, the sum will be 4 times the number. What is the 
number ? 


DEFINITIONS. 19 


17. One number is 3 of another; their sum is 156. What 
are the numbers ? 

1s. 7*@—52+112=390. Finda. 

19. Divide the number 99 into three parts so that the first 
shall be 2 times the second and 3 times the third. 


20. A man bought two houses for $ 4400, paying 10 times as 
much for one as for the other. What did each cost? 


21. A man paid $700 more for one house than for another; 
the cost of one being + of the cost of the other. What was the 
cost of each ? 

22. One seventh of a number exceeds } of it by 560. What 
is the number ? 

23. What number added to 2 of itself will make 1000 ? 

24. What number diminished by # of itself will make 60 ? 


25. What number increased by 4 and } of itself will make 


CHAPTER III. 
ADDITION AND SUBTRACTION. 


27. Arithmetical Addition. In elementary arithmetic, to 
add two numbers, 4 and 5 for example, is to find in the 
number system a number 9 by the process of counting, 
first 4 and then 5 more. ‘The number so found is called 
the swum, and the numbers added are called the addends. 
The addition of any number of addends furnishes only an 
extension of the above process of continuous counting. 
If we think of number represented as heretofore upon a 
scale, then addition may be represented as follows : 


5 A 5) B 6 bs 


Let us add 5, 3, and 6. 

First, we count 5 from O, which takes us to A; then 3 
more, which makes 8 and takes us to B; and finally 6 more, 
which makes 14 and takes us to P. In this illustration O 
is the zero point from which counting proceeds, and P is 
the terminal point. Hence, the sum represents the counted 
distance of the terminal point from zero. In practical 
addition elementary sums are remembered, and thus the 
actual counting is avoided. 


28. Algebraic Addition. We have already seen that 
ordinary algebra contains not only positive number (for- 
ward counting), but also negative number (backward 

20 


Oe 


ADDITION AND SUBTRACTION, 21 


counting). Part or all of the addends may therefore be 
negative. Hence, the definition of algebraic addition 
must be extended accordingly. 


Algebraic addition is the process of finding a number 
(sum) in the algebraic number system represented by the 
terminal point reached by the successive forward and back- 
ward countings indicated by the addends. 


Thus to add 4, 5, — 6, and 2 is to count.4, then 5 more, 
giving 9, then backward 6 to 3, then forward 2 to 5; 5 is 
the terminal point of the successive countings and is the 
sum of the given addends. 


On the diagram the counting is from O to A, A to B, 
Bto ©, Cto P. 
To add 4, — 6, — 38, and +2 is to count 4, then back- 


- ward 6 to — 2, then backward 3 more to — 5, and then 


forward 2 to —3. The terminal point of the successive 
countings is —3, which is therefore the sum of the addends 
given. Show this by a diagram. 


EXERCISES. 


Find as above the sum in each of the following, and illustrate 
by a diagram : 


1. 4, 5, —3. 6. —8, —9, —4. 

2. 6, —T7, —3. 7. 4, —8, +6, —5. 

3. —4, 5, —1. 8. —6, +9, —10, +8. 
4. 8, —9, +6, —4. 9. —4, —2, +8, +6. 
5, 


a es 10M Shee Omer 10 


Paya THE ESSENTIALS OF ALGEBRA. 


29. Monomials and Like Monomials. Algebraic expressions 
consisting of single terms are called monomials. 

Monomials containing the same literal parts, each literal 
part having the same exponent, are called like monomials. 

62, —4ab, 3 xy, are monomials. 4a, 2a,and — 3a are like mono- 
mials. To add 4a, 2a, and — 3a, is to count 4a, and then 2a to 6a, 


and then backward 8a to 8a. 38a is the terminal point of the suc- 
cessive countings, and is therefore the sum. 


EXERCISES. 


Add the following monomials: 
4b, 5b, —2b, —3b. 5. ll abe, —10 abe, —4 abe. 
30b, —6a*%b, —5a’bd. 6. 8a’, —52’, +727, —Oz-. 
—5ay, Tay, —2 ay. 7. —2a’y, +4a°y, ay, —d ay. 
6ax’, 3ax*, —10 aa’, +ax. 8 IYaxy, —Saxy, —Taxy, Lazy. 
9. 3Vay, —2Vay, +6Vay, —V2y. 
10. 4avz, +5avz, —9avz. 


Pew N eT 


30. The Commutative Law. We know that in arithmetic 
8+4=443. This is known as the Commutative Law, 
and means that the addends may be taken in any order. 
The law is applicable to any number of addends. This 
law holds in the extended number system of algebra. It 
is algebraically stated as follows: 


a+b=b+a. 


The following diagrams show the truth of this law: 


ADDITION AND SUBTRACTION. 23 


That a—6b=—6-+4a is shown as follows: 
O a B 
Y i =? 
B “0 O 
PRAMS iva Dare qa ee re Vig gi.) 
+c 


The distance from O to P is the same in each case. 


31. The Associative Law. We know that in ordinary 
arithmetic, in adding 2, 3, 4,5, we may, if we wish, first 
add 3 and 4, and add their sum to 2, and then add 5. 
We may, in fact, associate the addends in any manner we 
choose. ‘This is known as the Assoczative Law. ‘This law 
holds in the extended number system of algebra. It is 
algebraically stated as follows: 


a+tb+ec=a+(b+e). 


This law, like the Commutative Law, applies to both 
positive and negative numbers. 


32. Addition of Monomials. Let us find the value of the 
expression 
Sa— 2a +47 — (a+ 82. 


By applying the Commutative Law we may write this 
expression in the equivalent form 32+427+82—22—1 72. 
Now, by the Associative Law we may add all the positive 
numbers into the one sum of 152, and all the negative 
numbers into another sum of —92, and thus write the 
expression in the equivalent form 15% — 92, which we at 


24 THE ESSENTIALS OF ALGEBRA. 


once know to be 62. All this may be shown in the 
following scheme: 
8a—2r74+42e7-—T2r14+82 
=3r+4xv+8x—2x—T2x, by Commutative Law. 
= lbr—9za, by Associative Law. 
Sad by adding. 
The sum of a positive and negative number is equal to 


the difference of their absolute values with the sign of 
the greater prefixed; e.g. 6—4=+ 2, while 9-—13=—4. 


Rute. To add like monomials, add the positive terms, 
then add the negative terms ; to the difference of the absolute 
values of the two sums prefix the sign of the greater. 


EXERCISES. 


1. Add 3a, —5a2, —9a, +112, —32, +72. 


We may for convenience arrange the solution thus: 





+ 32 — 52 
lice — 92 
ie ae 3a 
2Zlix —17e% 

=172 

+ 4% 


It will be noticed that we have arranged in the first column 
all the positive nuinbers, and in the second all the negative 
numbers. This is a mere matter of convenience. The stu- 
dent should early accustom himself to pick out and unite the 
positive numbers mentally, and likewise the negative numbers, 
merely writing down the results. Unless the coefficients are 
very large, a little practice will enable the student to do this 
with accuracy and rapidity. 


ADDITION AND SUBTRACTION. 25 


Add 5ayz, —11 xyz, 18 xyz, —4 xyz, —13 xyz. 
Add 15 a’bc, 10 abc, —2a’bce, — 18 a’be. 

Add 1027, —52’, —2a’, 1627, —82’. 

. Add $ax, tax, —fax, —lLax, — az. 

Mads V2 e,.—b5V22, —8V 2a, —V2 2, 6V2 a. 
7. Add 4(a+6), —3(a+b), —4(a+)b), 2(a+0b). 


Algebraic expressions containing like quantities within a paren- 
thesis may be added as monomials. Thus, in Exercise 7, the quantity 
a + 6 is common to each addend, hence the sum is — (a + b). 


8. Add 5(@’+ 7’), —3(a’?+y"), T(a’+y’), —8(a’+y’). 
9. Add t(ax+ by+2), —3(ax+ by+2), (av+ by+ 2). 
lo. Add 3(@*—2y), —§@’*—2y), —4@’—2y). 

a1. Add 3(Vz+Vy), —8(Va+vVy), 6(Va+vy). 


oy? eT op? 2 
12. Add eet) Al eiiatimto) eae) 


33. Addition of Polynomials. An algebraic expression 
consisting of more than one term is a polynomial. 

If it has two terms, tt is called a binomial: and tf three, 
a trinomeal. 


A Aa Pw DW 


If we desire the sum of 4a + 6 and 3a — 46, we may indicate that 
sum thus: 4a +60+3a—4bD. 

Now, by the Commutative Law, we may write this 4a+3a+b—44, 
which by the Associative Law becomes 7 a—3b. 


However many polynomials we might have to add, the 
process would be an extension of the above. We have 
then the following rule for adding polynomials: 


RuuE. Unite the like terms of the various polynomials 
into sums, and connect these by their a signs to form 
the polynomial sum. 


26 THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 


1. Add 2ax4+5by+5cz, 4av—S5by, Tby—8 cz, 1lax— 
4 by — 6 cz. 


For convenience we may arrange these polynomials thus: 


Zax+3 by+5cz 
4 ax —5 by 
T by —8 cz 


11 aw — 4 by — 6 cz 
1Tax+ by—4cz 


It should be noticed that in the arrangement like terms have 
been placed in columns. 


2. Add 5a+5y+42, —3a—5y+z2, Tx—2y—z, 10%+y—2z. 

3. Add 5ax—6 by+cz, Sax+by+5cz, —4ax+3 by—8 cz, 
30 ax—3 by +3 ce. 

4. Add 42°+3y'?—4, —30°+27?—6, 2? —y? +7. 

5. Add 380° +4Y7+432?, —e?—47°4+222, 11 e—yY—-—2 

6. Add 4ay—S8 yz2+7 2a, 4yz—Tay, 3yz+420, —4ayt+ 
320 — 2 yz, — xy — 2H. 

7. Add 4Vay+3Vyz+V2zx, —2Vyz2—3V2x, —2Vay— 
BV y2 + -V 20. 

8. Add 8abec—4ayz+Tlmn, —4lmn+12ayz, 5 ayz— 
4abe+3lmn, —5 abe —T lmn. 


34. Identity. The sum of the addends is identically 
equal to the sum. 

Thus, 8a+4xe—52x7=22. 

Also, 2a+6+5a—4b—6a+4b=a+4+6. 


Since an identity is true for all values of the letters involved, we 
may make use of the identity existing between addends and sum to 


| ADDITION AND SUBTRACTION. Q7 


verify our results in addition. If in the first illustration above we 
make x = 1, it becomes 
34+4—5=2, or 2=2. 


If in the second illustration we put a=1 and b =1, it becomes 
24+14+5-4-64+4=1+41, or2=2. 


The use of 1 for each of the letters is convenient, but not necessary. 
We may use any value whatever. Thus, in the second illustration 


above, we may put Hee Sinviat Bele 6 
The identity then becomes 
6+5+415-20-—18+4+20=3+4+5, or 8=8. 


EXERCISES. 


Add the following quantities, and verify the results by sub- 
stituting particular values for the letters used: 

1. 35a+4)b), 5a—6b, Ta—4c, 5b+411¢. 
4%—3y, 5a+3y, —Ta+y, 3u—2y. 
2¢7—30, 50+40, 7T@7?—5B, —38a?— 02 
5a +4ay+3y’?, 2e°—B5ayt+6y’, 32’°—8y’, 8 ay. 
—dsaxr+4by+e, 5ax—6by—38c¢, aw—3by4+2e 
12%@—3y+2, 6a¢—4y4+72z, —8x—38y4+382. 
etyte2, sar+4y+522, —4e°—8y'?4+ 27. 
w—3y, be+4y4+3, vty, y—swe. 
8Va—4Vy, 5Va+8Vy, —6Va2+Vy. 

10. 4a? —5b?+65b, 3a? +203—40b. 

Nore. a? means Va; be means Vb. 

11. 5Va—6Vb+¢, —3Va+4vb—4e, 2Va—3V0 + 2c, 
N/a —38V/b4+3¢. 


12. Saxy+4 byz4+ 6czu, 2 byz —5axy+ 6czx, —4axy + 2 byz 
—Tczx, 2axy — byz — 7 cea. 


oO YO ah wD 


28 THE ESSENTIALS OF ALGEBRA, 


35. Subtraction. Subtraction is the process of finding 
one addend, when the other addend and the sum of the twe 
addends are given. 


The given addend is called the subtrahend, the given 
sum the minuend, and the addend which is to be found 
the remainder or difference. 

Referred to the number system, we may say that to 
subtract a from 6 is to find the amount and direction of 
counting necessary to pass from a to 6, both a and 6 begin- 
ning at the zero point. 

This is shown on our number system diagram as follows: 


b 


a. 


If P is the terminal point of the subtrahend a and Q 
that of the minuend 6, then the distance and direction 
from P to Q is the result of subtracting a from 6. 


36. Cases of Subtraction. A consideration of the defini- 
tion of subtraction gives us the following four cases: 


(1) If we subtract a positive number from a positive 
number, the remainder is the arithmetical difference of the 
absolute values, positive or negative, according as the 
absolute value of the subtrahend is less or greater than 
the absolute value of the minuend. 

For example: 


Ta—(+38a)=4a, Ta—(+10a)=— 38a. 


ADDITION AND SUBTRACTION, 29 


(2) If we subtract a negative number from a positive 
number, the remainder is the positive arithmetical sum of 
the absolute values. 


Forexample: 52—(—32)=8u. 


(3) If we subtract a positive number from a negative 
number, the remainder is the negative arithmetical sum of 
the absolute values. 


For example: —6ab—(+4ab)=-—10ad. 


(4) If we subtract a negative number from a negative 
number, the remainder is the positive or negative arith- 
metical difference of the absolute values according as the 
absolute value of the subtrahend is greater or less than 
the absolute value of the minuend. 

For example: 

—Ta—(—I1l2)=42, —138¢%—-(—-9xv)=— 4-2. 


The student should verify the truth of these cases by 
testing them on the number system diagram. 


EXERCISES. 


From 192*y subtract 11 ay. 
From 7ab subtract — dab. 
From —Q9ayz subtract 12 xyz. — 
From —17a subtract —15¢. 
From —142? subtract — 23 27, 
From 27 a’y subtract 43 a*y. 
From 157’ subtract 15 7’. 

From —7 aba subtract — 7 aba®. 
From 17 Vzy subtract — 12 Vay. 


See Se Sie ee Nac 


30 THE ESSENTIALS OF ALGEBRA. 


10. From 11(a+ 5) subtract 15 (a +4 b). 

11. From —6(e@+y) subtract 9(@ +7). 

12. From —11(a’+ 0’) subtract — 7 (a? + 6’). 

13. From 18 (ay + yz+ 2%) subtract — 7 (ay + yz + zz). 
14. From 23(aa + by +c) subtract 40 (ax + by +c). 
15. From —16(y’?+4 ax) subtract —21(y’+4 az). 


37. Subtraction of Monomials and Polynomials. In the 
cases considered in Section 86 it may be noticed that the 
subtraction of any number is equivalent to the adding of 
an equal opposite number. 

Illustrations : 


Ta—(+4a)= Ta+(—4a)=3a. 
Ta—(—4a)= Ta+(+4a)=11a. 
—Ta—(—4a)=—Ta+(+4a)=— 3a. 
—Ta—(+4a)=—Ta+(—4a)=— Ila. 


Hence, we have the following rule: 


Rute. To subtract one number from another number, 
change the sign of the subtrahend and proceed as in addition. 


For example: The problem, from 18 ax*y take — 5 aa’y is 
equivalent to this problem, to 18 aa’y add 5 ax*y. 

In practice the change in sign should always be made men- 
tally. 


Rue. To subtract one polynomial from another, change 
the sign of each term of the subtrahend and proceed as in 
addition. 

EXERCISES. 
n bk From 9e°+ 38ay—117°4+7 
Subtract 7a? —5ay—17y°+9 
2e°+8ay+ b6y—2 





ADDITION AND SUBTRACTION. ot 


In this we think of the 7 2? as negative, and add it to 92”, 
giving 2a. The —5ay is thought of as positive, and added 
to 3ay, giving + 8zy, and so on with the other terms of the 
subtrahend. 


2. From 9a? —11a?+5y subtract 14a°—7a’—8y. 

From 11 at — 15 b* — 13 ab? subtract 7 a* + 108*— 3 a7b*. 
From 4ay + 3a"? —11y’ subtract 7 vy —42?— 13 7°. 
From 18 ax —14 by + 11c subtract 5 ax + 19 by —5e. 
From 5 Va +8 Vy —134 subtract 8 Vx—5Vy+3a. 
7. From 192°+72—5 subtract 42° + 11a? —8. 


On See 


Arranging for subtraction, we have 
192 +Ta2—5 
4771 oe —§ 
15e¢ —1l1l¢?’4+Te+3 
In the minuend there is no term in x’, or we may say there 
is the term 0a. Hence the 112’ is to be subtracted from 0 2”, 
and of course gives a remainder of —112a*. Inthe subtrahend 
there is no term in 2, so there is nothing to subtract from 7 a, 
or the remainder is 7 a. 





8. From d2* —3 2° +-32-+6 subtract 6a* —32°+ 2? — 5. 
9. From 4 277’? +6ay?—3a*y+y' subtract 3 y'+4 ay—6 ay. 
10. From 3(a+6)—5(@#+y)+6c subtract 6(@+4+ y)— 
5 (a+b) —5d. 
11. From 14 -Vxv+13 y?—16 x subtract 15 y8+17 y—12 Vay. 
12. From 4az+5y7?—62z+13 subtract 21417 «z—15awu 
+11 y. 
13. From 18 4? — 272° + 42 yz subtract 18 yz — 36 y? +17 2’. 
14. From 36 a’ — 27 a®b —17 ab? + GO subtract 13 6°+ 17 ab? 
— 37 ad. 


15. From 137? +17 «wy — 5 subtract 18 y?—17 yz +32. 


O2 THE ESSENTIALS OF ALGEBRA. 


38. Identity. Subtraction may be expressed as 
Minuend — Subtrahend = Remainder. 


This is an identity and will, therefore, be true for any 
values of the letters involved. ‘This identical relation 
furnishes a convenient method for verifying the results 
of subtraction. 


For example: Tab—14a?+ 110? Minuend. 
4ab—11a2+4+ 19062 Subtrahend. 
8ab— 8a%*— 806% Remainder. 

If we put a =1, b =1, we get 
(PUES SO Beaten o 
4—1) 4°19 = 12 
eg oe ee: 


which shows our result to be true. 


EXERCISES. 


Perform the following subtractions and verify by putting 
the letters each equal to 1: 


1. From 9a?+ 11 ab—18a take 8 a?+14ab —9a. 

From 3a—4y+7 take «—2y+8. 

From 42° +57'+7 ay take 52’? —4ay4+3y?, 

From 14 aw +12 by +7 take —3 ax + 2 by—5. 

From 4(@+ y)+72—4 take 3@+y)—52—8. 
6. From 16(a’?+ ay +’) +2 —w’ 

take 12(a + ay+y?)-4245w". 
7. From 2(b?—4 ac) + 2+ 7? take 5(6?—4 ac) + 52?—6 7. 
8. From ax+ by+c take a'v+ dy +c’. 


Gis Ph WN 


ADDITION AND SUBTRACTION. 33 


9. From aa’ + 2 hay + by? +2 gx +2 fy 
take ax? +2 h'xy + b'y?+2g'e@+2 flyt+ cl. 
10. From 2+ 7’?+ aw + by take 32? +3 y+ le+ my. 
11. From aV2+bVy-+c take a Va2+ b'Vy +c’. 
12. From 5-V2? + 9? + 7-V/y? + 2 take 8-2? +? +8V y+ 2. 


39. Removal of Parentheses. If we recall the principles 
of addition and subtraction already developed, we can by 
means of them remove parentheses preceded by the + or 
— signs. 

For ~a+(6—c+d)=a+b—c¢+d, 
and a—-(6—e+d)=a—b+ec-—d. 

Hence, a parenthesis preceded by a + sign may be omitted 
without any change in the signs of the terms inclosed. A 
parenthesis preceded by a — sign may be omitted if the sign 
of each term within it is changed. 

For example: | 

ax+(3ab—4cy—3az)=axr+5ab—4 cy — 38 az. 
ax —(3ab—4cy—8az)=ax—38ab+4cey4+ daz. 


EXERCISES. 

Remove the parentheses in the following and unite like terms: 
1. 3a—(2a+4b—c). 2. 37—4y—(2e%+a—3y). 
3. 5a’?—(b6+3a’) —4b—(2a°—38 D). 

4. 3ax—[a?—3 ax + ay — (ay —5 aw +a’). 
Remove the [ ] first, and we have 
3 aa —a’?+3 ax —ay+ (ay —5 ax +a’). 
Now since the parenthesis is preceded by a + sign it may 


be omitted, and we have, when we unite the like terms, aa as 
the result. 


34 THE ESSENTIALS OF ALGEBRA. 


2a—8b—[7a—(4b—5a—b425)}. 

3av—5 by —[7 by + 3 aw — (5 ax —3 by). 
lla—(Ty—32)+5y—[8x—(4y+5a)]. 
5ae—-{Ta—[3e—(5a—2e%—a)]t. 

17a12e— fi5a—ll2+(8a—22)=—be— fat 

10. 2~— {6y+[6z2—(@—y+2)—22]+d3y}. 

In exercise 10 and the following put «=5, y =4, and z= 2, 


and find the value of the expressions. 


Ml. by—32+fa—y+22-—(8a42y—42)]. 

12. 32—{2a—[52—(4y—Tx—2y4+2)—3a]+2 yf. 
13. Qn+3y—(4y—z2+a) —[5e—(Ty4+82)]. 

14. xy — (y2— 2a —2 xy — 2 YZ). 


15. 38ay— {4e°—-y—[2Zay4+3a°—5y]—-CBx2’—-4y)}. 


oO OND gw 





40. Insertion of Parentheses. Many times it is just as 
important to insert parentheses in an algebraical expres- 
sion as it is to remove them. Evidently any number of 
terms may be inclosed in a parenthesis without change if 
the parenthesis is preceded by a + sign. Any number 
of terms may be inclosed in a parenthesis preceded by a 
— sign, if the sign of every term so inclosed is changed. 

For example: 


ax + by —cz+ab=axr-+ (by —cze+ab), 
ax + by — ce+ab=ax—(— by + cz —ab). 


The number in the parenthesis is thought of as one quan- 
tity, and hence may be considered as one term. ‘This gives 
an extension in meaning to the word term. In the illus- 
trations just given the expressions on the left have four 
terms, but those on the right have only two terms. 


ADDITION AND SUBTRACTION. oO 


EXERCISES. 
1. 3a?4+4ab—3c+d. Inclose the last three terms in a 
parenthesis preceded by —. 


2. ay+ by —cy—w«y. Inclose the last three terms in a 
parenthesis preceded by —. 


In each of the next four exercises inclose all the terms con- 
taining @ in a parenthesis preceded by —. 
3. 59—38xu+au—4abe. 4. Ty+d3z2+20—384u—5ypx. 
5. 15—5ax+112—16 a2-+ 27a. 
6. 27 y—30y 4+ 22ay—13b4 4 212%—50a4+16b. 
In the following exercises put all the terms containing a in 
a parenthesis preceded by —, and all the terms containing y 
in a parenthesis preceded by —. 
7. 15—8a+4y—S5av+T7ay+3ab—5 by. 
Rearranging, we have 
15—8a—5ax+3ab+4y4+T7 vy — 5 by. 
Then 15—(8a+5ax%—3 ab) —(—4y—T ay +5 by). 
8. 252+5y—824+7 ax—11 by+cew—38ac+11. 
9. b°—ab+cy—2ac?—4b*y4+1la—4y. 
10. 3%+2a—4y+5e2—6ab+3yz2—5au+11 by. 


41. Adding and Subtracting with regard to a Named Letter. 
 Ta+5a—6a might be written (7 +5—6)a. So if we 
had 7 az + 5 bx — 6 ex, we might write it (Ta +56 — 6e)z. 
In this, z has been chosen as the element of likeness or 
the denomination in the three terms. We have added the 
coefficients, but in this case the coefficients are unlike and 
we can only indicate the addition. We can subtract 5 aa? 
from 7 ba? if we consider 2? as the element of likeness. 
The remainder is (7 6 — 5a) 2%. 


36 THE ESSENTIALS OF ALGEBRA, 


EXERCISES. 
Unite with respect to x, 5ax +7 ay —3 xz. 
Unite with respect to a, 3ax+4ay— daz. 
Unite with respect to k, 4k +3 yk —12 zk. 
Unite with respect to x and y, 4ax+6 ay —5 ba +12 cy. 


YE ee Naa 


Unite with respect to x and y, 
3ax +4 by — (2 ax + 12 by). 
6. Unite with respect to 2’, 
ba + 16.07 — 12 be a 
7. Unite with respect to a, y, 2, 
3a+4y4+32—(5%—8y—T2). 
(83—5)a+(44+8)y+(847)2=24412y7+4102. 
8. Unite with respect to # and y, ? 
da—2y4+7—(2a4+3y+4). 
9. Unite with respect to a, y, and z, 
—5a4+8y+724+ 6a—4y+4+2). 
10. Unite with respect to the powers of a, 
3802 —52?—8e4+ 72? —162+4 202°—212+6. 
11. Unite with respect to the powers of w and y, 
aa” + by’ + ca + dy +e — (a'x’ + bly’? + clu + d'y + e'). 
12. Unite with respect to 2’, xy, and 7’, 
4a? —3 ay +129? — (20? —4 ay +8 y?) + 16 ay. 
13. Unite with respect to powers of k, 
12¢+16yk—64%4122k4+ 807k? +12%°4+ 2%. 
14. Unite with respect to a’, y’, 2°, and ay, 
aa” + 2 hay + by? + cz? — (a'a? + 2 h'ay — b!y? + c'2?). 


ADDITION AND SUBTRACTION. Si 


MISCELLANEOUS EXERCISES. 


1. Aand B have $550; A has $100 more than twice as 
much as B. How much has each? 

2. Add 11 aa’?+138 ba’y+9 cay’?—5 dy’, 4 ba’y—38 aa’? +16 dy’, 
31 ba’y — 24 cay”, and 10 dy’ —8 cay? + ax’. 

3. Remove the parentheses and simplify, 

13 2 —{8y—5a— (8e4+T7y)—2Zyi— (12%+8y422). 

4. Unite with respect to a and y’, 

aw — 5 a? + 34? — by? + ca? — dy’. 

5. From 5a°— 6 a’b —7 ab?+11 0° take 100?'—3a'°+ 5 ab? 
— 12a’), 

6. By means of a diagram show that the sum of 8 and 
—10 is —2. 

7. By means of a diagram show that —10 taken from 8 
leaves 18. 


8. By means of a diagram show that 8 taken from — 10 
leaves — 18. 


9. With «=4 and y=—5, find the value of 
Sy—j{sa—[4y4+2xe—(6e—3y—52a)}}. 
10. With the same values of w and y find the value of 
{8e—[3y+2e—(5y—382)]} (8e-—3y—2). 

11. A father’s age is 10 years more than 3 times his son’s 
age; the sum of their ages is 82 years. Find the age of each. 

12. Add, 8y¥y4+3ay+112, Ty—138ay—212, —14y+ 
10 ay +1227, and 2y—5 ay — 2 2. 
Verify your result by putting y= 2 and x=1. 


13. From ie ape ob 
take 8a+ Sab—-1006+11 


Verify your result by putting a=1 and b=2. 





38 THE ESSENTIALS OF ALGEBRA. 


14. Add 8(a+b) —16(#—y) + 11Vaae, 5(@— y) + 2Vax— 
7(a+b), and —38(a+b) +11(@—y) —14Vaz. (Regard the 
quantities in each parenthesis as a single term.) 

15. From 8(a?— 0b’) —17(«#+y) —11vV2'+/ 
take 7 (a? — b*) — 20(@+ y) + A/a? + y? 

16. Show by a diagram that the sum of 8, —6, —3, +2, 
and —5 is —4. 

17. Unite with respect to a and 2, 

oa+8a—4a—Tx—ba+ bat cx — de. 

18. 8%e+3y—(«+y)+2yz. Inclose all the terms after the 
first in a parenthesis preceded by a — sign. 


19. A farmer exchanged a bushels of wheat at 6 cents a bushel, 
and ¢ bushels of corn at d cents a bushel, for a mowing machine 
costing e dollars, and for calves at f dollars each. How many 
calves did he get ? 


20. John, James, and Henry together have $216. John 
has one half as much as James, and Henry has as much as 
both John and James. How much has each ? 


21. Simplify 8a°4+3ay—4y+2ay+3y—Tw+4y7— 
Q2ay—Ty+5e?+3ay—4y? +102” 
22. Simplify 17-—134+44a?— #—(S6y+ 21¢+52— 11). 
23. Ifa=—1, y=2, and z= 4, find the valine on 
wye — fa? —[y? + 22 — (22 —3ay—4y)] + 227}. 
24. With the same values of a, y, and 2, find the value of 
Byz+{8a—[4y+2— (Say +4 yz) —8ayz]— 34h. 
25. Add 3(a+y)— 4(a?—b)+ 3V/—P 
—Vety +11 —b) —16VP SF 
4(ea+y)— 8(@—d)— 4/75 
B(a+ty)— 8(a@—b)— IVP —F 


CHAPTER IV. 
MULTIPLICATION AND DIVISION. 
MULTIPLICATION. 


42. Integral Multiplication. ‘To multiply 7 by 8 may be 
taken to mean that 7 is to be added 8 times. 


TH+74+74+74+74+7+74+7=56. 


In such an operation 7 is the multiplicand, 8 is the multt- 
plier, and 56 is the product. Similarly, to multiply a by 6 
is to find the sum of a added 6 times: 


ata+ta+--(b times) = ab. 


In this, a is the multiphcand, 6 the multiplier, and ab 
(vead, a multiplied by 6) is the product. It is clear that 
this view of multiplication has no meaning when the mul- 
tiplier 6 is fractional or negative. We can not add a one 
half times, or three fourths times, or three and a half 
times; neither can we add a negative four times to obtain 
the product of a by —4. All this shows that if we are to 
have multiplication of algebraic numbers, we must extend 
our definition. No change is made in the meaning of 
multiplicand, multiplier, and product. 


43. Multiplication Defined. Multiplication rs doing to the 
multiplicand what has been done to unity to produce the 
multiplier. 

39 


40 THE ESSENTIALS OF ALGEBRA. 


For example. To multiply 6 by 5 is to do to 6 what has 
been done to unity to produce 5. Unity has been added five 
times (1+1+1+1-+1=5) to make 5, hence we must add 6 
five times to get the product 6x5. Or, we may say that 
unity has been taken five times to produce 5, hence we must 
take 6 five times to produce 6 x 5. The new definition is seen 
to agree with the arithmetical notion of multiplying one integer 
by another. ‘To multiply 6 by 2 is to do to 6 what has been 
done to unity to produce 2. Unity has been divided into 3 
equal parts and 2 of them taken to produce 2. Hence to 
multiply 6 by 2 we must divide 6 into 3 equal parts and take 
2 of the parts. It is thus seen that the new definition includes 
multiplication of fractions. The application of the definition 
may be seen by considering the following simple concrete 
problem. 

A water tank with a capacity of 1000 gallons contains at 
the present time 600 gallons. Let us consider the following 
four cases: 


(1) If it have a supply pipe carrying 10 gallons an hour, 
how much water will be poured into the tank in the next 
84 hours ? 


Evidently the amount poured in is 10 x 81. We must do to 10 
what has been done to 1 to make 8}. To produce 84, 1 has been 
added eight times, then separated into 2 equal parts and one of the 
parts added. Hence we must add 10 eight times, giving 80, then 
separate 10 into two equal parts of 5 each and add the 5 to 80, giving 
85. Hence the amount poured in is 85 gallons. 


(2) With the same rate of flow, how many gallons must be 
added to find the amount in the tank 6 hours before the pres- 
ent time ? 


In this case the 6 hours is negative, as it is the opposite of the 
time used in the preceding problem, which we considered positive. 
Our problem now is to multiply 10 by — 6. We must do to 10 what 


MULTIPLICATION AND DIVISION. 41 


has been done to 1 to produce — 6. — 6 is produced by subtracting 
1 six times. Hence to multiply 10 by — 6, we must subtract 10 six 
_ times, which gives — 60. Hence we must add — 60 gallons to 600 
gallons to get the contents 6 hours before the present time. 


(3) A discharge pipe is flowing at the rate of 5 gallons per 
hour. How many gallons must be added to give the contents 
10 hours from now ? 


Since the inflow rate was positive, the outflow rate is negative. 
Hence, our problem now is to multiply — 5 by 10. We must do to — 5 
what has been done to 1 to produce 10; that is we must add — 5 ten 
times, which gives — 50. So — 50 gallons is the amount to be added. 


(4) The discharge pipe has been running at the rate of 6 
gallons an hour for a number of hours. How many gallons 
must be added to give the contents of the tank 8 hours ago? 

In this case both the 6 and 8 are negative, and the problem is to 
multiply — 6 by — 8. We must do to — 6 what has been done to 1 to 
produce 8; that is, we must subtract —6 eight times, which gives 


+48. So we must add 48 gallons to get the contents of the tank 8 
hours ago. 


The results of the four cases may be arranged thus: 
10 x 8§ = + 85, 
10 x —6= — 60, 
—5x10= — 50, 
(—6) x (—8)= +4 48. 
Tf we should make our reasoning perfectly general by letting 


the rate of flow be a and the number of hours be 8, then the 
four cases would take this form: 


oxo =-+ab, 
ax (—b) = —ab, 
—axb = — ab, 


(—a) x (— 6) = + ab. 


49 THE ESSENTIALS OF ALGEBRA. 


44. Law of Signs in Multiplication. From the definition 
and the above considerations, we see that the product is 
4- if the multiplier and multiplicand have like signs, and 
it is — if they have unlike signs. 


In multiplication, like signs in multiplicand and multiplier 
give a positive product and unlike signs give a minus product. 


45. Continued Products. Products produced by three or 
more multiplications are called continued products. 


If 6 boys buy 5 oranges each at 3 cents apiece, the total cost of the 
oranges is 3x 5 x 6 cents. If 8 groups of 6 boys each should buy 
oranges as above, the total cost of the oranges is 3 x 5 x 6 x 8 cents. 
If instead of using arithmetical number we should use algebraic num- 
ber, and say that there were d groups, c boys in a group, and each 
boy bought 6 oranges at a cents apiece, then the total cost of the 
oranges would be a x b x ¢ x d = abcd cents. 


46. Factors. As in arithmetic, the numbers multiplied 
together are called the factors of the product. 


a, 6, and e are factors of abe. 

5, a, 6, and e are factors of 5 abe. 

In the latter case, 5 is considered a numerical multiplier 
or coefficient, and is usually so designated, instead of being 
called a factor. 


EXERCISES. 


Point out the factors and numerical multipliers in the follow- 
irg products: 


1. abex. 4. 11(22)(3y)z. 
2. Sabdz. 5. (52) (4a) (82). 
3.17 xX 9(8 2) =o) x92. 6. (Sy) (42) (2 x y). 


MULTIPLICATION AND DIVISION. 43 


47. Signs of Continued Products. 
(—3)x2x8 =—6x3 =— 18, 
(—3)x(-—2)x3 =+6x3 =+18. 
(—8)x(—2)x(—3)=+6x(—3)=— 18. 

If we use algebraic instead of arithmetical number, the above 

may be written: 

(—a)xbxe =(—ab)xe = — abe. 
(—a)x(—)b)xe =(+ab)xe = + abe. 
(—a)x(—6)x(—e¢)=(+4 abd) x (— ¢) =— abe. 

We see that one negative factor produces a negative 


product, two negative factors a positive product, and three 
negative factors a negative product. Hence, 


Products resulting from an even number of negative fac- 
tors are +, those resulting from an odd number of negative 
factors are —. 

EXERCISES. 
Give the signs of the following products: 

1. (—2a) x (5a)(—2). 
(—a)(—b)(—0)(-@ 
a(—b)(—0)d(—e) f(—9) (— A). 
'(— a) (— b) (y*) (— 82). 
17 (—5) (— 3) (— 2) (eyz)(—2)(—8). 
48. Commutative Law of Factors. In multiplying to- 


gether numbers in arithmetic, the factors may be taken in 
any order without changing the product. 


Hence |. SS OD: 
eee a OX OX OS ox OS 8K Sx 2 
meee x. ah xD Kh 2 = 30. 


Oe SS 


44 THE ESSENTIALS. OF ALGEBRA. 


If we construct a rectangle of length 5 and width 8, the 
area is 5x38. If we construct another rectangle of length 
3 and width 5, the area is 8 x 5. 3 


qQ 


These two rectangles are readily seen to be equal, for 
the second one is merely the first one placed on end. 


Hence, 5x8=8.x 5. 
The reasoning would be exactly similar if we should use 
the general rectangle whose length is a and width 6, and 
the other general rectangle whose length is 6 and width a. 
In this case our conclusion is 
ake Osed. 

The product is the same whatever the order of the factors. 

This is called the Commutative Law of Factors. 


This law holds for all algebraic numbers. 
ax (— b)=(— b)a =— (ba) = — (ab) = — ab; 
or, since —b=(—1)4, 
ax(—b)=ax(—1) xb=(—1lab =— ab. 
EXERCISES. 
1. Commute the factors of 3 ay. 
SRY = 3 YX SUSY = yd = YOM eee 


2. Commute the factors of — abe. (—1 is a numerical 
multiplier.) 


MULTIPLICATION AND DIVISION. 45 


3. Commute the factors of ab(@+y). (The quantity in 
parenthesis is a factor.) 
ab(a + y)=a(@+y)b=(a+y)ab=(a+y)ba=b(x+y)a=ba(x+y). 
4. Commute the factors of (a + b)c(x + 9). 
5. Commute the factors of (a + b)(¢ + d)(e+/). 


49. Associative Law of Factors. If 6 boys buy 4 oranges 
each at 5 cents apiece, the total cost is 8 cents x 4 x 6. 
We may think of this as each boy paying out 3 cents x 4 
and the 6 boys as paying out (3 cents x4) x6; or we may 
think of total number of oranges bought, 4 x 6, and then of 
the cost of these as 38 cents x (4 x 6); or finally we may 
think of the six boys as buying 1 orange each, at a cost 
(8 cents x6), and then the total cost of 4 oranges each is 
(5 cents x6) x4. Hence, we have 


Ween (5 X4)x 6=3 x (4x 6)=—(8 x 6) x4. 
If instead of the arithmetical numbers, 3, 4, and 6, we use 
the algebraic numbers a, 6, and c, we have 

eee a XO) xX C= a xX(b xX ¢)=— (a x ¢)x O. 

The above is an illustration of the Associative Law of 
Factors, which may be stated as follows : 

The factors of a product may be grouped in any order. 
Ge exo —(ab) X (cd) =a(bx e)d=a(bxexa). 
8x(—8) x (—4) x(—2)=1[8x (—3) ]} x {(-4) 

x (—2)}= etc. 

50. Distributive Law of Factors. ‘The multiplication of 
4465 by 7 may be indicated thus: (4+5) x7. The opera- 
tion may be carried out in either of the following ways : 

(4-5) x l= 9 x T= 63. 
(44+5)7 =4x74+5xT=28+4 35= 63. 


46 THE ESSENTIALS OF ALGEBRA. 


In the first instance the 4 and 5 have been combined 
into 9, and the product of 9 by 7 taken. In the second 
instance the 7 has been distributed as a multipler of the 
terms 4 and 5 of the multiplicand, and the sum of the 
separate products taken. If these arithmetical numbers 
4,5, and 7 be replaced by the algebraic numbers a, 6, 
and e, we have 

(a+b)xe=axc+bxc=act+be. 

The fact expressed by the algebraic identity 
(a+b) xe=ac + be 

is known as the Distributive Law. Since the multiplier 
and multiplicand are commutative, 

ex(a+b)=(a+ 56) x c=ace + be. 
_ It may be noticed that the Distributive Law harmonizes, 
as it should, with the definition of multiplication. To 
multiply a+6 by eis todo to a+6 what has been done 
to 1 to produce ec. This would certainly mean that we 
are to do the same thing to a and 6 and add the results. 

In distributing factors the law of signs must be observed. 

(a—b)x5ce=5ae— 5be. 
(— 2a? + 6) x(—38¢)=6 a@e—3 Be. 


EXERCISES. 


Distribute the following factors: 

- (a+b—c)x 2d. 

(2 eB i. te) C8 oh, 

. 2at+ty—3s2)2a). 

. 12 a —5 yy? + 15 ay) (4d). 

. (15 wy — 15 yz + 12 az)(— 2 ad). 


ao P&P WO N 


MULTIPLICATION AND DIVISION. 47 


(ab —5 0? + 38¢ —a)j(dxy). 

(ax — ba + cay)(— 3 mm). 

(4 lx +5 my — 3 nz)(2 ad). 

(— 6a + 5a’b — 38 ab? + b*)(— 3 2’). 
10. (aa + bx + c)(— 2 yz). 


oO NO 


51. Index Law of Factors. Ina continued product, abed, 
any two or more of the factors may become equal. ‘To 
indicate the product when two or more factors have 
become equal, a convenient notation has been devised. If 
6 should become equal to a in the above product, we would 

have aacd, and it would be written a2ed. The small 2 to the 
right of the a and slightly elevated is called the exponent. 
If 6 and ¢ each equal a, we have aaad =a'd. If 6b, c, and 
d each equal a, we have aaaa=a'*. It should be noticed 
that the exponents 2, 3, and * indicate in these cases the 
number of times @ occurs in the respective products. 


a-a=a’, read a square. 
a-a-a=a’, read a cube. 
a:-a:a-a=da', read a fourth power. 
a-a-a--tomas=a", read a exponent m, or a mth. 
It should be noted that a= al. 
W=A+Aa+a-a-a. 
Wxrr=Aa+-a-a+xe'a. 
Write out the equivalents of 
1) dz?= 
(2) 5y = 
(3) Tad-d-a:a°4:4x-¢x= 
(4) a2?2(38 = 


©) 2-2-2.2-2-¢7-9-e-e-rs 


48 THE ESSENTIALS OF ALGEBRA. 


@X@W=a-4-4X4:4=4-0:0:0-:4=0=0", 
a" Xa"=(A-a+a- tOMAMS)(4+a:a- tO NAS) 
=(a-a-a-+- to (m+n) a’s) 
sete aes 
The result CR 
is the Index Law for positive integral exponents. It is read 
a with exponent m multiplied by a with exponent n equals 
a with exponent m+n. 
ge. = ot? = 75, 
y' z af” as Nek on y”. 


EXERCISES. 


a? - at. P= (a*- a?)(a*- a) =a" a, 
8 


ey P= 


(ety) @ty-(@+y)P= 
(a+ b)*- (a+8)%-(@+y)?-@+y)= 
BT y+z-yt2)- (a—b)-@—b= 


SOOO eae) aaa ee a Oo ear 
oy) 
bo 
ou 
oo 
tc 
_ 


_ 
9 


52. The Multiplication of Any Number of Monomials. 

RULES: 

(1) Write the product of the numerical coefficients. 

(2) Attach the literal factors of the product, observing the 
Index Law for repeated factors. 

(3) Prefix the proper sign, determined by the number of 
negative factors, + if an even number, — tf an odd number. 


These rules result from observing the laws already developed. 


MULTIPLICATION AND DIVISION. 49 


Multiply together 3 ab, 4 ac, 5 bc, — 3 abe. 

The indicated result is 

3ab x 4ac x 5be x (—3abe) 
=3.4-5-(—3)a-a-a-b-b-b-c-c-e (by Commutative Law) 


=-—180a-a-a-b-b-b-c-c-c (by multiplying numerical 
factors and observing law of signs) 


= — 180 a*b*c3 (by Index Law) 


The above has been developed by an application of the laws, but 
the result is in exact accord with the rules on page 48. 


(2a) x (— 8 ab) (— 2 bc) = + 12 abc. 
EXERCISES. 


Multiply together the following monomials: 


1. 3ax,4ab, —S5aay. 5. —4lx’, —3 By’, —8 aty’. 

2. 5a’x, 3 b’y, 2 a®a°y"*. 6. 2abc, 3 b’c®, — a*be’. 

3. —4 mn, 2mn*, —S5m‘n’a?. 7. 3(a+b)*, 4y(a+b)’, 5(a+b)* 
4. —axyz, Day’, —4 aye’. 8. 8 au’, 5(a+b)*, —2a%(a+b)?. 


53. The Multiplication of a Polynomial by a Monomial. 
RULES: 


(1) Distribute the monomial as a multiplier of each term 
of the multiplicand. 


(2) Connect the results by the proper algebraic signs as 
determined by the law of signs. 


By the Distributive Law 
(a+b) xec=ae+ be. 


6 can be any number, as d+e; then (a+ b)c=ae + be 
becomes 


(a+d+e)e=ac+(d+e)c=ac+dc+ee. 
In general, 


(at+b+e4+d+-.-)ms=am+bm+em+dm-+-:-. 


50 


L; 


THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 


(5e+3y)x 2a=1020+6 ya=10 ar+ 6 ay. 


It is customary to write the letters of a product in alphabetic 


order. 
Law. 


ae ee Le 


YH » eS 
QO N HF © 


14, 


15. 
16. 


17. 


This can always be done by an application of the Commutative 


(8a+4b)x2a=6a'4+8 ab. 

(83a?—4ab)x4ab= 

Multiply 3 a°%—2a2?+3ay by 5 axy. 

Multiply —3 a+ 4la’?—Sdlay by —2 Pa’. 

(ax +3 ay—5 bey) x 8 axy = 3 way +9 aay? — 15 abcay’, 
Multiply a?— ab + ab?— B® by —a%b®. 

Multiply 3 at— 4a’ — 6 ab? +7 ab® by ab’. 

Multiply a’?+6?+a’+y+2 by 6 aba”. 

Multiply 5ab —12cd—3,fh by —6be. 

Multiply —4 a’yz+7 ay’z—3 xyz? by — 5 ayz. 


. Multiply 3 a’b? +7 b’c? —14 ca? by 8 a*b°c’. 
. Multiply 3(@@+y)?+4(a+b)? by 2(@+y)(a+5). 


B(w +9)? + 4(a +d) 
2(a+ y)(a +d) 
6a@+ya+b)+8@+y)(at oy 
Multiply 
(7+ yP—3a(e+yP?+4y(e+y’) by 5 ay(a’+ 7’). 
Multiply (a—b)*—38 a(a—b)?+7 y(a—by* by 2 x’y(a—b)’. 
Multiply 
3(@+yyP+2a(e+y)/P?—38H(e+y) by —ab(a+y) 
Multiply 


5(a? + y)? — 38 a(a? + y)? —T atb(a? + y) by —4(a? + y)*. 


-_ MULTIPLICATION AND DIVISION. 51 


18. Multiply 
10(a? + 3 b?)* — 6(a? + 3 6)? +12(a@2 +3 b’) by —5(a?4 3 0?) 
19. Multiply (a+6+c)?—3(a+b+c) by 5(a+b+¢)*. 
20. Multiply 
(au’ + bx +c)’ + 4(aa? + ba +c)? by — 5(aa’ + ba +e). 
5%. Meaning of 0 x a and a x 0. 
(6—c)a=ab—ae. 
If 6=e, this becomes 
(¢ —¢)a=ac— ae. 
But e—c=0 and ac—ac=0. 
Hence, Nie AUP 
By the Commutative Law, 


Oxa=ax0=0. 


55. Multiplication of a Polynomial by a Polynomial. RULE. 
To multiply a polynomial by a polynomial multiply each 
term of the multiplicand by every term of the multiplier and 
take the algebraic sum of the results. 


This is a direct consequence of the Distributive Law. 
Find the product of (a+6) by (@+y). 
Let x+y be replaced by e. 


Then (a+ 6) x(#+y)=(at6) x c=ae + be. 


But c=r+y, 
and so (at+b)(e@+y)=ac+be=a(rt+y)t+bet+y) 
=axr+ ay + bx + by. 


This product consists of the sum of the products of each 
term of the multiplicand by each term of the multiplier. 


52 THE ESSENTIALS OF ALGEBRA. 


If three polynomials are to be multiplied together, two 
of them must be multiplied as above and their product 
multiplied by the third. 

Thus 

(a+b)(a+ y)(r+8)=(ax + ay + bx + by) + 8) 
(by multiplying factors one and two together). 
(ax + ay + bx + by) (7 + 8)= are + ary + bra + bry 
+ asx + asy + bsx + bsy. 


EXERCISES. 
Find the product of the following polynomials: 
1. GBa+bd)(a+b)=8¢0C+ab4+8ab4+0?=38a°?+4ab4 BD. 
The distribution may be arranged conveniently as follows: 
oa+ 0 
a+ 0 
30+ ab (the product of (8a-+ b) by a) 
dab +b’ (the product of (8a +6) by 6) 
3a? +4ab+ 6° (the total product) 
This is called long multiplication, and should only be used 


when the distribution can not easily be made in the straight 
line form illustrated above. 


2. (@+y)(@—y) =e 4 ay —ay— Y= ve — yy’ 


3. (2e%+y)(2Qea—-y) =4e-y. 

4. (4a4+38y)(4a—3y). 8. (x+ 5) (@— 2). 

5. (au + by) (ax — by). 9. (y+ 4)(y’—3). 

6. («+ 4)(@+4+ 2). 10. (ax? +b)(x+c). 
7. (©—2)(@+ 38). 11. (8%+5) (4% + 2). 


ee 


12. (4a°+4+ 32°45 a—6) (82 —2a°?—324 2). 


MULTIPLICATION AND DIVISION. 53 


Multiplicand and multiplier are both arranged according to 
the powers of 2; that is, the exponents of a decrease uniformly 
from left to right. The arrangement will be just as good if we 
reverse both factors and have the powers of @ increase from 
left to right. For multiplication we may arrange the work as 
follows: 


4a? +3a°+ 5a — 6 
oa —2a— 3a-+ 2 
120°+92°415e'—182? (product of the multiplicand by 32°) 


—8x2°— 62*—102?+122’ (by — 2 2”) 
—  —129'— 9o3—150°+182 (by —3 2) 
| oe + 6274+102—12 (by 2) 





oe ae oe — 292+ 3274+ 28e¢—12 (the total product) 


The orderly arrangement according to powers of « merely 
insures that the terms of the product will come in an orderly 
way, thus making it easier to arrange like terms in columns 
ready for adding. When arranged in this way the highest 
power in the product appears first, and is the product of the 
two highest terms in the factors. 


—- 13. (4@—5a?+3a—6)(8—@+2a). 


When these factors are arranged according to powers of a, 


we have 
(—5a?+4a°4+3a—6)(—a’?+2a+43). 


Now multiply as in Exercise 12. 
—14. (—6y+4+77’—12)(4—2y+4+7’). 
When arranged, this exercise becomes 
et YOY 12) GF + 0 yf! 2 y-+4). 


Observe that in the multiplier the term 7? does not occur ; 
in the arrangement according to y we write 0’. 


54 THE ESSENTIALS OF ALGEBRA. 


15. (b¢—54+60?—3b+26*) (hh —264 2). 
- 16. (@—Ta+120 + 6)(@—38a—5S). 

~17. (82°+4+120°—10%+4 4)(—2—5+4 82). 

~18. (5a°y? —6 xy +12 2°y? — 4) (8 —42%y? + yx’). 
-19. (42—-4+462—52)(—42z+43). 

~ 20. (80?+4at+a+5a’°—4)(at—a@+a?—a+1). 
al, (2084 3a2—42—1)(8a+44). 











SOLUTIONS. 
(1) | (2) 
2¢ + S3e— 47 — 1 2+ 3— 4-71 
ox + 4 o+ 4 
6at+ 9a®—120°— 3a 64) 
$a 42 a 16 eee § +12 =16—4 
Gat t17e+ Oa—192_4 6417 +0 


In solution (1) the multiplication is carried out in the usual 
way, while in solution (2) merely the coefficients are used. 
In the answer on the right the #’s should be inserted, beginning 
with a, The method used on the right is called multiplying 
by detached coefficients. It is a device which saves time by the 
omission of all letters. 

In using detached coefficients the following directions should 
be observed : 

(1) The multiplicand and multiplier must be arranged 
according to the same letter. 

(2) 0 must be used as the coefficient of every power of the 
letter of arrangement which does not occur. 

(3) The letter of arrangement is inserted in the product by 
beginning at the left with a power equal to the sum of the 
highest powers in the multiplicand and multiplier, and decreas- 
ing uniformly to the right. 

The following exercise illustrates this: 


MULTIPLICATION AND DIVISION. 50 


22. (3a°+42°4+50°4 7x41) x (52°—32). 
See 4 4 Ot he 
oo — 3 
1o6e-0 +20 +25. +35 + 5 
“— 9 -—"0 —12 —15 —21 —3 
P. 


foe Ue lia + 25a 4+ 23 ae — 102 — 21a — Sa 


The highest powers of « in the two factors are 5 and 3, so 
the product begins with 2%. 
In the following exercises use detached coefficients. 


— 


—23. (at—5a?+40—384+2a)(80—5a+2). 

mated eo oa — 2a-+7)(a’+4a° —da’?+ 2a — 7). 
~ 25. (ay? —4 277? + 6 ay — 5) (a*y? + 4 wy? — 6 xy + 5). 

— 26. (32% —Tae+4a—5)(22 —32+4 4). 

— 27. (Ty —5y+3y—4A4y—8y+1). 

— 28. (a*b®?— 6 ab + 7)(5 ab — 4 a°D"). 

— 29. (4a‘b'— 7 — 6a’b? + 3 ab)(ab —5 + a’d’). 

- 30. (807+ 4a— 5) (a’?+2+41)(@’—382x+3). 

56. The Identity in Multiplication, The multiplicand 
multiplied by the multiplier is identically equal to the 
product. 

Multiplicand x Multiplier = Product. 


(a+b) xec=ac+t+ be. 


Let a=b=c=1. 
Meee Teed. vad sc dy 
PCT Weer pad Me oe 


aay, 
A — we 


56 THE ESSENTIALS OF ALGEBRA. 


This furnishes a convenient method of verifying the results 
in multiplication, If in Exercise 22 above we put «=1, we 
have 


(8+4+547+4+1) x (6—3)=15411425+423—-10—21—3. 


D0. eee 
40=40. 


This merely verifies the coefficients. In order to verify the 
exponents, some value other than 1 would have to be sub- 
stituted. In case more than one letter occurs, numerical values 
must be given to each. 


(a+y)(@—y)=e7—y’. 


Let Ce DRT 99) ees 
Then (142) —2)=1?— 2% 
(3)(—1) =1-—-4 
— > = — oe 
we EXERCISES. 


Perform the following multiplications and verify by means 
of the identity : 
-1. (+ ab)\(2a+43D). - 5. (5a°—S8ay42y’)(2a+3y). 
2. (4a4+6b)2Qa—4b). -6. (8%+4a)(8a%—4a). 
— 3. (724+ 3a+41)(a?—a#+1).- 7. (6 ab —7 cd) (5 ab 4 T cd). 
~ 4 (5ab +3 cd)(2ab—4cd). —8. (6a°y—4ay’+y") (8 cy—Ty’). 


57. Involution. In multiplication when the factors are 
alike, the operation is called involution, and the result a 


Owe? 
P aA-a-a=a’, 


Qs a? - 9? =" 07 )8 == nd 1a 


(a® . a® . a®) = (a°)® = aaa aaa aaa, 





MULTIPLICATION AND DIVISION. 54 
58. Meaning of (a’”)”. 
(a”)” means a” -a™. a”... to n factors. 


Since each of the n factors, a”, contains m a’s, there 
are in the product mn a’s. 
But a” means a-a-a-a-- to mn factors. 


Hence, (ae er art, 
More generally, (a7b')" = a™"b™. 
The exponent n is distributive as to factors within the 
p 
parenthesis. oS 5 ae 
ay")? = vy”. 


The exponent is not distributive as to terms within 
the parenthesis. 


@+yP=@t+NetYNEety-. 
(7 + y)? is not equal to 2? + y?. 


The difference between the following forms should be 
noted : 


Ae na a’. a” = hn"). 
(a)? aid ae. (a™" — gy, 
EXERCISES. 


Remove the ( ) and simplify: 
. (a7)* (a®)? = a’ - ab =a". 


(2) (@yyt 


3? . (3), 

(ae ys (aay, 
-(@t+y)y - (8°)? « (27)? (a?) (a)? 
(42) ty’) @+yy-@+y) 
Roa = 2°. 26 = 218, 10. (ab’c)* + (a®be®)'. 


Oo » © N HB 
(ace st oO) 


08 THE ESSENTIALS OF ALGEBRA. 


DIVISION. 


59. Division Defined. — Division is the process of finding 
one number, when the product of two numbers and one of 
them are given. The given product is the dividend, the given 
number the divisor, and the required number the quotient. 

Division is the inverse of multiplication, the dividend 
corresponding to the product, the divisor to the multiplier, 
and the quotient to the multiplicand. 

Since axb= ab, 

ab+b=a. 


60. Law of Signs in Division. 

From multiplication we have 
(+ a)(+ 6) =+ ab. 
(+ a)(— 6) =— ab. 
(—a)(+ 6) =— ab. 
(—a)(—b6)=+ ab. 

From the definition of division it follows that 
+ab+(+b)=+a. 
—ab+(—b)=+a. 
—ab+(+b)=—a. 
+ab+-(—b)=—a. 

Like signs in dividend and divisor give a positive quotient, 

and unlike signs give a negative quotient. 


Gl. Index Law. We already know that 


q'” x q” a qmtn, 


Hence, qntn Pen qr ee a” — qQmtn-n, 
Suppose mM+n= Dp, 
then aP +a" = ap-", 


More generally, aPb6%e" + a"bsct = q?-"b4-ser-t, 


MULTIPLICATION AND DIVISION. 59 


For the present it is understood that the exponents of 
the factors of the dividend are not less than the exponents 
of the corresponding factors of the divisor. 


The above considerations show that the exponents of the 
Factors of the divisor are subtracted from the exponents of 
the like factors of the dividend in order to obtain the expo- 
nents of the factors of the quotient. 


Illustrations: 

eet iat 3. 

(2) aP6-+ a3h5 = q0-3)6-5 = ib, 
(3) atb°® + a®bect = abrct. 


4) @+y%e+uy'+ ty E+uy=(@+yetw)® 


62. Meaning of Z and a, 
a 


We know that ax0=0. 
Hence, We 0. 
a 
(oii Agee EAH 
Hence, ert 
7 ks 
But, by Index Law, a” = a" = a, 
om 
Now by Axiom 1, i ea 


Any quantity with an exponent 0 ts equal to 1. 


63. Division of One Monomial by Another Monomial. 
RULES: 

(1) Divide the numerical coefficients as in arithmetic. 

(2) Attach the literal part determined by the Index Law. 
(3) Prefix the proper sign determined by the law of signs. 


Peer Onyer <= 3 72 —=:3.0%2, since 4 = 1. 


Fa 


60 THE ESSENTIALS OF ALGEBRA, 


a EXERCISES. 
Divide: 


21 a’a®y? by 7 ara’y. 

72 atay’2" by 24 a®y’2™ 

108.a7y*2? by 36 aty’2?, 

81 a°b®cd? by 27 a°b*d. 

144 a(x + y)’ by 48 (a+ yy 

— 63 (a — b)’ (@ — y)' by 21 (a—b) *(w—y)?. 

AB a*y?(a® + y?)? by —5 y (a? + y?) 

3° atayz by 3? a’a*y. 

— T*BaPyt(z— ax) by — T?Pa'(z— a). 

3f- F(a — @)*(b — y)* by 3? - 5°(a — w)7(b — y)*. 


A SS OY eh ae eee 


= 
9 


64. Division of a Polynomial by a Monomial. From the 
Distributive Law of Factors we know that 
(a+b+ec) Xk = ak + bk + ck. 
Hence, (ak+bk+ckh)+k=a+b+e, 


which shows that # is distributed as a divisor to every 
term of the dividend. 


Rute. Divide each term of the polynomial by the mono- 
mial and add the results. 


10 a2? + 5 ax?y — 20 a8x*y? 
5 ax 


=2av+y—4ay?. 
EXERCISES. 
1. (12 a®y* — 8 atay?’ — 4 a’a*y*) + 4 a®y’. 


2. (Qabcx* — 18 a7b*a* + 27 a®bc’ax) + 9 aba. 
3. (380 a*y%z + 25 a*ytz? — 35 a*y'2”) + 5 a*y’e. 


MULTIPLICATION AND DIVISION. 61 


4. (@+ y+ 5@+y)'—aw@t+y), (Regard x + y as a term.) 
(@+y) 
5. [4(a — b)* — 5a(a— 6)? + 11 ay (a — b)’] + (a — )*. 
6. Divide 5 ay(a+y¥)*— 10 ay? (@ + y)? — 15 ay’ (@ + y) by 
5 wy (w+ y/)*. 
7. Divide 11 a*b?(a?+ a?) + 22 a% (2 + a’)? — 33 ab? (a? + a’)? 
by 11 ab(@’+ a’). 
8. Divide 
T xyz (a? — b*)? + 21 ay’2? (a? — b?)" by —T7 ayz (a? — 0), 
9. Divide 
24 a*b’c (aw + b)* — 36 a’b*c’ (ax + b)’ by — 12 (a*b’c) (aw + b)*. 
10. Divide 
—33 (aa’+ba+c)?+44 at (aa’?+ba+c) by 11 (aa’?+ba+e)’. 


65. Division of a Polynomial by a Polynomial. The divi- 
dend and divisor should be arranged in descending or 
ascending powers of some common leading letter. This 
gives a quotient arranged with respect to the same letter. 

The first term of the quotient is found by dividing the 
first term of the dividend by the first term of the divisor. 
The process is illustrated in the following solutions : 

(1) Divide 

2 —a*— 11234 1622-22-38 by 22-4243. 
2—4¢+3)2°—2t—11234+162—-22—3(0°4+38 22?-22-1 


—4e'+ 323 = 23(22?—4 2443) 
8a*—14 23 +16 22-2 7—3=1st partial div. 
32'—1243+ 922 =3 022-4443) 


—223 +7a2?9-2¢—3=2d partial div. 

—203 +8227-6a2 =—22(2?—42+3) 
—z*+4x2—38=3d partial div. 
—2°4+427—38=—-—1 (#?—42+4+3) 


62 THE ESSENTIALS OF ALGEBRA. 


‘This scheme of division is merely a separation of the 
dividend into parts. In the example just solved we have 
separated 2° — af — 112° + 162? — 2x —3 into these parts: 
(—4 a'43 23) + (3 at—12 2?+9 27) 4+ (—2 23 +8 2-62) 

+(—a?+4x—5). 
Now, regarding the dividend in this separated form, we 
have the division thus: 

ge —4et+38e8  8at— 1228 + a? — 2224+ 822-62 

w—47+4+38 7 4g lB we—4ateg 
a oe ope teat == 7 oe ad e 
(2) Divide at — 16 by 2+. 
z+ 2)a* — 163 —2¢74+42-8 








+e 








g4+ 223 
— 27° 
— 222 — 427° 
+ 472 
—8r—16 
— 82—16 


Divisions such as the above, which terminate without 
any remainder, are called exact. 


EXERCISES. 
Divide: 
1. «t—a®—92°+154%—12 by «—3. 
2. #—5e—38x2+15 by «—5. 
3. 60° +7e2—1824+5 by 24+ 5. 
4. 2e*—9a*?+172a?— 14a by wv —2e. 
5. Sat—6a0°+2a°4+14a—21 by a’— 2a438. 


MULTIPLICATION AND DIVISION. 63 


a*+4a° +6070? +4 ab?+0' by a+2ab4+ 0% 
wv’ —a® by a? — a’. 
b’c®’ — a® by de® 4+ at. 
m—3m'+3m?—1 by m?—1. 
10. a” + 2a"b" +b" by a® +b" 
11. a” —b™ by a®— 0" 
12. a’"—b™ by a®— b”. 
13. af —130?+472?—312+4 by 2 —6a+1. 
14. «*—120°+ 54a?—108e%+481 by a’?—6249. 
15. ab? —Sa’*b'cd + 5 aberd? — c'd® by ab — cd. 
16. 12 a°y§ — 17 vty? +10 vy? —3 by 4a°y?— 3. 
17. x°—10a*+40 2—80 a +4+80e—382 by a—4a44. 
1s. 7+15¢—212°+18a?—4a* by T—62+4 2", 
19. (a+ b)?—5(a+b)+4 by (a+b) —1. 
20. («+y)?+7(e+y)—18 by (w+ y) 49. 
21. (a+ 2)?—(a+a)—42 by (a+) —1. 
22. (m+n)? —-T(m-+n)—44 by (m+n)+4. 
23. (a+b)?—x* by (a+b) —«@. 
24. («+y)>—(a+b)* by (@+y)— (a+D). | 
G66. Detached Coefficients. When the dividend and divisor 
are arranged in descending powers of some common letter, 
the quotient is also thus arranged. We may then perform 
the division by the use of the coefficients only. 
(1) Divide 2*4+23— 327+ Tx—6 by 22-2742. 
1—1+2)14+1-—-84+7-614+2-8 


op Orga 








(h=qheky 
higeds 9 ha We 9 
ya yee 

—3+3—-—6 


64 THE ESSENTIALS OF ALGEBRA, 


Since at-+22=.2?, we know that the quotient begins 
with 27, and is 77+22—83. 

In the use of detached coefficients all powers of the 
letters from the highest to the lowest power must be 
present in both divisor and dividend. If any powers are 
absent, they must be inserted with zero coefficients. 


(2) Divide 2—8 by x—2. 


In the dividend neither z* nor 2 appears. We insert 
them, writing the dividend 23+ 02?+ 02-8. 


1—2)14+0+4+0-—8014244 





1—2 v4+2x+4, quotient. 
2+ 0 
2—4 
4—8 
4—8 
EXERCISES. 


Divide, solving by detached coefficients : 
w—5a’?+4 by «—1; by «+2. 
at—Toe4+110°+T¢e%—12 by x—-1; by «—4. 
e'—132°+ 36 by w+a—6. 

at —18a°y? —175y* by wv — 257. 

2m* —17 m®n + 31 mn? — 23 mn? +12 nt by 2m —Sn. 
at — 256 by a? +16. 

b§— 729 by b?— 9. 

y®— 4096 by x? +8. 

a+a*t+1 by at—a?+1. 

16a*+4a?+1 by 4¢+4+2a+1. 

. +4 by +2e42. 

. 1864150°-170°—3 by 50?—46+438. 


SO Cee a Oa UIE SE 


Hoe 
Nb fF Oo 


MULTIPLICATION AND DIVISION, 65 


13. a°— 0 by a? +2 0 + 2ab?+ 5b. 
14. 2+y by at — aty + a*y? — ay + y'. 
15. 10 a*— 48 a*b + 26 a*b? + 240d? by —5a?+40ab4+30% 


67. Inexact and Continued Division. 
(1) Divide 27+ 1 by x+1. 
: x+1)22+1(7—1 
w+ ex 
—z+l1 
—x—l 
2 
In this example there is a remainder of 2, and the 
division is inexact. In such examples the division should 
continue until the largest exponent of the remainder is 
less than the largest exponent of the divisor. 
(2) Divide 1+ 2? by 1 +z. 
1l+e2)14+21—24+2a2—228 


bee oes 
— pt 7 
aT ee 7 
2 x 
2 a2 +- 273 
— 273 
— 248 2 at 


2D at 


dol 


This division may end with two terms of the quotient 
and the remainder 22”, or with three terms and the re- 
mainder — 223, or with four terms and the remainder 2 2+. 
Evidently the division might be continued to any number 
of terms desired. When inexact division takes this form, 
it is called continued division. 


66 THE ESSENTIALS OF ALGEBRA, 


EXERCISES. 
Find quotients and remainders: 
1. Divide 2? —72?+11x%—7 by +3. 
2. Divide + 72? —8%+413 by a? +3a—2, 
Divide a—11@+ 21 by a +5. 
4. Divide bt—146°+4 11 by 0? +9. 
Divide a’a’? + 9 a’a’? — 7 by ax + 3. 


sa 


oO 


In the next five exercises continue the division to four terms: 
Divide 1 +a by 1 —@. 

Divide 2+35a%—42° by 2— 2. 

Divide 3 —6%+4 82? by 8+ 52a. 

Divide 1 —17x%+4 132 — 82° by 1— 1624 527, 
10. Divide 10 — 20a + 25 a*?— 31a by 2—5a. 


© OND 


68. The Identity in Division. 
Dividend + Divisor = Quotient. 
(—1)+(@—-l=274+er41. 


This is true for all values of z. 
Let z= 2, and we have 


(23 —1)+(2-1)=274241, 


(8 —1)+1 =142+1, 
T+1 = 7, 
ff = 7. 
In this case, should we make «= 1, we get 0+ 0 on the 
left of the sign =. 0-+0 is indeterminate. In using the 


identity to verify divisions, avoid substitutions that will 
produce this form. 


MULTIPLICATION AND DIVISION. 67 


EXERCISES. 


Divide and verify by substituting particular values: 


1. 62 —177 + 241-16 by 2P—3/1+4+4. 
2. y —1lby y—1. 

3. 2 +1 by «+1. 

4. a—120?+ 48a — 64 by a’ —8a+ 16. 
5. 625 — 5002 4 1502? — 2022+ 2* by 5 —2. 
6. 625 —2* by 254 2. 

7. y'—dy — 154 by 7? — 14. 

8. a‘b* + 4a°b? — 117 by a’b’ + 18. 

9. 2° —y° by a’? — yx’. 

10. at—4a?—34a°+76a+105 by a—T. 


REVIEW EXERCISES. a 


1. Find the value of 6 2? — 4 ay +12 77, whenv=4,y=—1. 

2. Find the value of wz’? — 64y°+82—3 ayz, when «=0, 
Boe = OD, 

3. From 16 a? —4 7° + 12 2 — 14 a’y + 3 ay’ subtract 12 y+ 
8a°+4 ay? —10 ay 4+ 11 2’. 

4. Remove parentheses and unite like terms: 

16 —§12a+[46—3c]+8—[8a—38(4—2))]}. 

5. Remove parentheses and unite like terms: 
—3a’+4[ay—ax(8a—4y)—3y(4ea4+2y)]—fa?+3(ay—y’)t. 

6. Unite terms in a, y, z: - 

ax + by + cz—4(— a'a + b'y —c') +3 (lx + my + nz). 


7. Simplify «—4y—[z—y—(@+y—z)], and find value “ 
Wie Ga yez = 1. 


68 THE ESSENTIALS OF ALGEBRA. 


8. Simplify 4(a—5 {b— ct) —[8 b+ }2b—(ce—a)}], and a 
find value when a=1, b=2, c=3. 


9. Find value of Pr RTE lcd 6) | ze + V2? ey — vine 
when wes) aye) ee, 


3 Va? + aval ee 


c+Y—z2 






10. Find value of 
when ¢=4, y=), 2= 6. 
11. Remove parentheses and simplify: 
a(y—2z) +y(z—2%) +2(@—y). 
12. Remove parentheses and simplify: , 
a? (y —2) + y? (2-2) +0? (2@—Y). 
13. Multiply #+a+41 by #—a+1. 
14. Multiply #’+7?+1—a—y-— xy by Petar 2: 
15. Multiply 2’ — 47? by a + ay’. 
16. Multiply 12a¢'—-3 03+ 102? -—5a2+4 by 322-2? +52 
— 4, 


17. Find the value of #—4a°+3a—5, (1) when x=2; 
(2) when «= —1; (8) when «=0. 

18. Find the remainder after dividing #—42a?+3a—5 
(1) by «—2; (2) by. «+1; (8) by =. 


Note that the remainders are the same as the results found in 
Exercise 17. 


19. Divide 2 a’ — 3 a‘b — 6 a°b + 18 a’b? — 6 ab® by 2a—3 b. 
20. Divide 3 a*+140°+92+ 2 by 2?+5a+1. 

21. Divide 2 a’°+ab—ac—30?’—4be—c by 2a+3b+¢e. 

22. Divide a —(a+b)x+ ab by x—a. 

. Divide a’ —(a + 6 + c) a’ +(ab + be + ca) « — abe by x—a. 
. Divide #2—-U4+m+n)2+ (m+ mn+iln)z—lmn_ by 





ee Pe (Ltn) z+ln 


oe 





MULTIPLICATION AND DIVISION. 69 


. Multiply together («+ a") and (@ + a). 

. Multiply 2” — y" by a +y". 

. Multiply 30"%4 5a"+7 by 2a"—4a—3., 

e Divide 3.47% +13 21415 a 4+ 9a by a+ Bar. 
. Divide xt” + ya — yma" — y™™ 4 a" y" by a+ y”. 


Multiply a?+0° by a—b, and divide the product by a+0. 


. Multiply 3a?44a'—22" by 2x?—32'+2". 
. Divide a” —b*”" by a"— bd”. 
. Multiply 2? 4+ 3a"t!—5a* by a ?—2a +a". 


Divide a” — by aim  ag3my™ gry” aeny” + y, 


. Divide a®— b%" by a”— b*. 


Divide (a + b)”" —«* by (a+ b)" +2”. 


meeieioe (ye + y)*—1 by (7+ y)*—1. 
(Vax +b)? + 4? by Vaa+b+y. 
. Simplify @’—ayt+y)(@t+ty@e-yy+@e+y). 


5(a® + a’b + ab? + b*) x 4(a—b)’?+ [2 (a? —b") x 10(a? +") ]. 


. CHAPTER V. 
IMPORTANT IDENTITIES. 


Many expressions in algebra appear in standard or type 
forms. When this is the case, multiplications and divi- 
sions can be performed mentally by. remembering certain 
identities. 


69. Multiplication Identities. 
1. The Product of Two Binomials with a Common Term. 
This form is given by 
, (x + a)(x + 6) =x°?+ (a+ 6)x+ab, 
or (a+ x)(6+x)=ab+(a+b)x+x°*. 
(1) (2+ 10)(w+5)=274+ (10+ 5)z + 50 
=27+ 152+ 50. 


To find the product of «+10 and x + 5 it is only necessary 
to see that a is 10 and } is 5 and make these substitutions. 


(2) (e—7)(«@+8) =a°+(—7+8)a—56 
=27’+a— 56. 

In this a is —7, and D is 8. 

(3) (542) (11 —a) =55+4 (11 —5)e—2, 

In this exercise it is necessary to note the signs of the 2’s. 

(4) (8e+y—5)(8e+y4+7)=B6e+y)?+2(8a+4+y) -385. 


In this exercise the common part is 3@+y. 
70 


IMPORTANT IDENTITIES. 71 


EXAMPLES. 


Write out the products of the following: 


1. (w+ 10)(x — 2). 6. (aw —11)(ax + 6). 

2. (3%+6)8e+1). 7. (3 xy — 5)(3 xy — 6). 
3. (Ta—d)(Ta+A4). 8. (42a°+7)(4a°— 5d). 

4. (564+2a)(54+6a). 9. (3—42y)(5 + 6 xy). 
5. (2y—6)(2y+7). 10. (a+ b—6)(a+b+5). 


ll. (€+60+4+7)(a+ b—8). 
12. (©+2ab—3)(x+2ab+7). 
13. j4—(@+2y)[ [5+ (@+2y)}. 
14. [(a+))?—4a][(a+b)?+7 a]. 
15. [Sab —(a—y)’|[5ab + (@— y)?]. 
2. The Square of a Binomial Sum. 
If, in the identity 
(e+a)(e#+b)=224+(a+b)x+ab, 
we let = tl, 
it becomes (x +a)(«@4+a)=274+(at+a)zr + aa. 
(x +a)?=x°+ 2ax+a’. 
The square of the sum of two quantities is equal to the 


sum of their squares increased by twice their product. 


(Set+4yP=OrP?+2O6xndy+Gy” 


=25 a+ 40 xy + 16 y?. 
EXERCISES. 
Write out the results in the following: 
1. (~@+y). 2. (2%+a)* 3. (34%+4b)% 
4. [(@+y+aP=(et+yy’+2aa+y+o. 
5. [yt+t(a+b)]. 7. [((a—3)+5 yf. 
6. [37+ (2a+c)}*. 8. [(a+b0)+(+y)}. 


re THE ESSENTIALS OF ALGEBRA. 


3. The Square of a Binomial Difference. 
If, in the identity | 
(x#+a)?=2742ar+4 a’, 
we change a to —a, it becomes 
(x —a)?=x°—2ax+a’. 


The square of the difference of two quantities ts equal ta 
the sum of their squares diminished by twice their product. 


(Za—y)y=(2a)?-2Cayt+y 
=4a*—4dayt+y*. 
EXERCISES. 


Write out the results in the following: 


1. (a—«@)*, 5. [(a+b) —ay]’. 

2. a—y)* 6. [(Ba+y)—ab]. 

3. (xy —4 b)*. 7. [(@+y)—(a+b)f. 
4. (3 a’— by)’. 8. [((2a—y)—dsa’y/*. 


4. The Product of the Sum and Difference of Two Quan- 


teties. 

If, in the identity 

(«+a)\(e#+b)=2?+ (a+b)r+ ab, 
we let b=—Aa4, 
it becomes (#@+a)(x—a) =2°+ (a—a)xu—aa. 
(x + a) (x — a)= x°—a’. 

The product of the sum and difference of two quantities 

2s equal to the difference of their squares. 


(1) (Tx—8y)\(T e438 y)=(T2)?— (8 y)?= 49 22— 9 of, 
(2) (ax+by—c)(ar+ by + ¢)=(ar 4+ by)? — & 


+ 


IMPORTANT IDENTITIES. 13 


EXERCISES, 

Write out the results in the following: 

1. (@—y)(u+y). 2. (3a—b)(Ba+b). 

3. (4x4 ab)(4 x%— ab). 
ne: [(a+#)+a][(a+2)—a]=(a+aP—V=0+2 aa+a’—@ 

= 2 aw +2". 

. [@+2y)—#][(@+2y) +2]. 
. (a? 4+ 2% 4+16)(a? + 2 a — 16). 
- [@+b)—-@+y[@+o)+e+y]. 
. (ax? + ba + ¢)(ax? + ba — c). 


NO Oo 


o 


5. The Square of a Polynomial. 
If, in the identity 
(e+a)2=2°+2ar+ 
we put a=Y +2, 
it becomes (a@+y+2)?=274+2e(y+2)4+(y+2)%. 
(xty4+zpexrt+y+ 2274 2x4 2xz+ 2yz. 
This may easily be extended to include the square of a 


polynomial of any number of terms, the result being that 


The square of a polynomial equals the sum of the squares 
of its terms increased by twice the product of each term by 
every other term. : 

(1) @t+y—avSr+y +a’ +2ay+22(—a)+2y(—a) 

HErt+yt+art 2ay—2av—2Zay. 
(2) (82+2a—4b) 
= (3 a)?+ (2.a)?+(—4b)?+2(32)(2 a) +2(3 7)(—4b) 
+ 2(2.a)(— 4b) 
=O +4¢°+ 1607 + 12 ar— 24 be —16 ab: 


74 THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 


Write out the results in the following: 


1. (a—av—yy’. 6. (ec —y+t+a+b)* 

2. (8x—y+d)?. 7. (m—n—p—q)’. 

3. (2a4+3b—y)’. 8. (2a+y—sz+a)’. 
4 (3a—5a—2y)* 9. (Ww—ab+2xe—s3y). 
5. (2+5a—4))*. 10. (2x —y? + ay — a)’. 
6. The Product of Three Binomials. 


By actual multiplication, it is found that 
(x +a) (x+6)(x+e)=x°4+ (a+6+4+e)x? 
+ (ab + bc + ca) x + abe. 

The product is arranged according to the powers of the 
common letter x The coefficient of 2? is the algebrate 
sum of the second terms of the binomials, the coefficient 
of 2 is the algebraic sum of their products in pairs, and the 
term free from z is the product of the second terms of the 
binomials. 

(1) @+1)(@+ 2)(@+3) 

=e+(14+2+43)e?+(1x241x 342 KS) ah ee 
sagt 627-11 oe G, 
(2) @+2)@—3)(a +4) 
=a + (2—3+44)a? + [2(—3) +2 ‘i 3)(4) Je 
+ 2(—38)(4) 
eo Boe 10 ods 


EXERCISES. 
Write out the results in the following: 
1. (w+ c)(x + d)(a + e). 3. (6+ 2)(6—1)(6 +38). 
+ 2. (atajy(aty(ate). 4 (y—3)(y+2)(y—1). 


IMPORTANT IDENTITIES. T5 


5. (m+5)(m — 4)(m — 3). 7. (a — d)(a — 3)(a? + 8). 
6. (ay +2)(ay—T)(ay 1). & (4) + 11)(y'—7. 
9 (@+y+5)(@+ty+3)@+y4+2). 
10. (8a+y—2)(8ea+y—4)(8x+y+4+6). 
ll. (aw+b+42)(av+b-+4 8) (ax+b—5). 
12. (ax + ba + 4) (aa? + bx — 2) (aa? + bx —1). 


£ T. The Cube of a Binomial. 
If, in the identity 
(e+a)(@+b)(@+c)=2+(a+b4e)2 
+ (ab + be + ca)x + abe, 
we put 6 and ¢ each equal to a, it becomes 
(7+a)(#+a)(e#+a)=22+ (a+a+a)z? 
+ (aa +aa+aa)zx+ aaa, 
or (xa)'=x°480xr43a'%x+a° 
=x*+a°+3ax(x+a). 
If, in the above identity, we change a to — a, we have 
(x —a)=x°—3ax?+3a°x— a? 
= x°—a®—3ax(x—a). 
The cube of a binomial is the cube of the first term plus 
three times the algebraic product of the square of the first 
term and the second term, plus three times the algebraic 


product of the first term and the square of the second term, 
plus the algebraic cube of the second term. 


A) @+2yfP=e+3@)'2y+38@2 y+ y) 


=0'4+6 ay+12 ay’?+8 7’. 
Ba) 2 by ==(2 2)°+-3(2 a)*(—3 b) +-8(2 a)(—3 by 
+(—3 b)? 


=8 a'— 36 a*b+ 54 ab?—27 0°. 


76 THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 


Write out the results in the following: 


1. (2a +50). 9. (4ab—5y)’. 
2, (a—3d)*. 10. (5 48 =). 
3. (8244) 2 
11. [(@+y)—2)%. 

4. (2xa—5y)* . 

a 12. [(w—y) +a]®. 
5. (3 a@ + ab)”. 13. [2(¢@+y)—3a)* 
6. (2 mn — pq). \ 14. [ (aa a b) { Ny 
7, SCD Seu 1s 15. [((a+b)+(#+y)]* 
8. (6y—1)y. 16. [(ax +b) — (ca+d) 


8. The Binomial Theorem. | 
(a + 6)"=a"+ na" + nin 2) = q?26? ata ie —*) 5 ve 2 a°5: 


f+ seeeeee tL nab-! a 6". 

The above identity is known as the Binomial Theorem. 
A general proof for it will not be given. For the present 
we will limit the exponent n to integral values. 

If we multiply both sides of the identity 

(a+b6)=a?+ 3 a7b+3 ab?+ 6? by a+ 6, 
we have (a+0)t=at+4 ab + 6 al? + 4 ab? + Ot. 

If in the binomial theorem we make n = 4, we have 


4-3 4-3-2 7 es es | 
By at ah eee eee 33 
(a+b)t=at+4a cA (a fy + Tone aa WT 


=a'+ 4 ab + 6 ab? + 4 ab? + 6, 


54 





a result which agrees with that found by multiplying. 


IMPORTANT IDENTITIES. T7 


If we multiply both sides of the identity 
(a+ b)t=at+4a% + 6 a0? +4 ab? + Ot by a+, 
we have 
(a+ b)>=a° +5 ad + 10 a30? +10 a7? + 5 nee iP 
If, in the binomial theorem, we make n = 5, we have 


(a+b)S=a5+ 5 ath + a : 





azb2 + - 


= 05+ 5 a!6 +10 a362+ 10 7004.5 “BA. 
a result which agrees with that found by multiplying. 
_ The identity 
(a+ 6)"=a"+ na” 6 ail aed ar a8 mn OS 2’) q” 68 


as Suk dss - nao + b* 
is often called the binomial expansion. 
The following Laws should be observed in regard to 
the Hxponents and Coefficients of the successive terms of 
the binomial expansion. 


(1) Law of Hxponents. The sum of the exponents of 
a and 6 in any term is always n; the leading letter a ap- 
pears in the first term with the exponent n which decreases 
by unity in each succeeding term; the letter 6 appears in 
the second term with the exponent 1 which increases by 
unity in each succeeding term to the last term 6”. 


(2) Law of Coefficients. If any term be taken, the 
coefficient of the next succeeding term is obtained by 
multiplying the coefficient of the given term by the ex- 
ponent of the leading letter a, and dividing this product 
by the number of the given term in the series. 


Tpaee THE ESSENTIALS OF ALGEBRA. 


Thus, in 
(a+ bys at + 4a + 6.00? +4 ado +O 
the coefficient 6 in the third term is obtained by taking the 
product of 4, the coefficient of the preceding term, by 
the exponent of a, 3, giving 4 x 8, and dividing by 2, the 
number of the term 4 a4 in the series. Hence, the coeffi- 
cient of the third term is 4x 3+2=6. 


Pascal’s Triangle. The coefficients of the terms in the 
expansion of 
(a+ b)!, (a+ 6)’, (a+b)%, (a+ b)4, ete., 


may be arranged in a table forming what has been called 
Pascal’s Triangle. ‘The arrangement follows. 


Coefficients of (a+6)!are 1 1 


Coefficients of (a+6)?are 1 2 1 
Coefficients of (a+ 6)? are 1 3 8 1 
Coefficients of (a+6)*are 1 4 6 4 1 
Coefficients of (a+6)® are 1 5 10 10 5 1 


etc. 


Each number appears as the sum of the number im- 
mediately above and the one to the left. Thus, the first 
10 is the sum of 6 and 4; the second 10 is the sum of 
4 and 6; the last 5 is the sum of 1 and 4. By this simple 
arrangement the binomial coefficients for any power of 
a+6 may be easily written out, provided we know the 
coefficients of the expansion of a+6 for a power one 
lower. Knowing 


Lieb LOTR ecm 


IMPORTANT IDENTITIES. 79 


to be the coefficients for (a+ 0)°, the coefficients for the 


sixth power of a+ 6 are 


1-6) ee Z0 eB 1 
(1) (a+2b)=at+4a%(2b)+6a7(2b)? +4a(2b)*>+ (25) 
= 0°48 ob - 24 0°b* ++ 32 ab? 16 bt 
2) @2—y)s (22)? +5(2 a) —y) +10(2 2) — y)? 
+ 10(2 2)" —y)? +5 (2 @)(—y)*+(— yy 
= 32 2° — 80 aty + 80 wy? — 40 a?y? + 10 ay* — 2. 


EXERCISES. 


1. Extend Pascal’s triangle to include the expansion of 


(a+ b)”. 


Write out the following expansions: 


6 btbriniie 8 

2. (a+b)*% (Use Paseal’s triangle.) i (« 1 s) 

3. (x+y) 9. (8y—22)*. ~ 

4. (m+n)*. 10. (2a +b)’. LG: , a cet 
ae 17. e+ Y)+o/*. 

: ep ll. (a@—3y)’. : 

: ° 4 2 : A » 6 18. [ (a+b) —2x]* 

6. (y+ 1)”. 12. (212+ a)* 19. [Sa—(a—y)]’. 

7. (24+ a)%. 13. (w —2y)". Wee 22 

8. (p— gq). 14, (a? — ¥)°. é Sey 

EXERCISES. 


By comparison with types write down the following products : 


a 


. (@+5)(@+3). 
(x + 10)(x— 2). 

. («—5)(a@+ 8). 

. (ax + 3)(ax + 5) 


&® BD N 


5. (vy + a)(ay — 6). 


6 
7 
8 


6. (4a+2)(4a—5). 


. (a? + 5) (a? — 10). 
. (w+a43)\(a@+a—7). 


wa 


80 THE ESSENTIALS OF ALGEBRA. 


ey 


10. 
is ip 
12. 


13. 


ey) 





(3 + by)(4 + by). 14. (Sx+y)(Bux—y). 
(7? +324 2)(@’+3a—5).. 15. (Qe—4y)\(2a+4y). 
Baty)(B«a—2y). 16. (2a+y+8)2x2+y—8). 


(a+b+12)(a+b—6). 
y 17. (a+ y?— 4)(a? + y? +4). 
25+ h)\(o—5—¥ . 
2/\ 2 18. (av + by + c)(ax+ by — ¢). 
19. (@’+1+42)(#?+1—2). 
20. (ay + yz +2@ —a)(ay + yz +244). 


Perform the operations indicated : 


21. (wt+y+z)’. 27. (a+ y")?—4 ary”. 

22. (8%+2y+2). 28. (v«+y+z2+w)’ 

23. (aw+by+c)*. 29. («+3)(~+5)(@+6). 

24. (x +3)?—122¢. 30. («+3)(a—3)(x+5)(a —5). 
25. (w+3y—4)?+16(e+3 y). 31. («+3 a)(2?—3ax+9 a’). 


36. 
Si: 
38. 
39. 
40. 
41. 
42. 
43. 


3xe+2 y+6)—24(3 2+4+2 7), 32. (~+35 yy)’. 
33. («+ y)?—S vy(a@+y). 
34. (a? +0°+C—ab—be—ca)(a+b+ ce). 
35. (2° 4+4y4+1—2ay—2y—a2)(a+2y+1). 


Show the truth of the following identities: 

(w+ a)? — (@— a)’ = 4 a2. 

(x + a)? — 4 ax = (a — a)’. 

(P+ ay +yja-—ysv—y¥. 

(P—wa +Pjeryav’t y. 

(a + ay + xy? + 9°) (a — y) = at — of. 

(2? — ay + ay? — y?)(x + y) = at — yf. 

(a? + wy +o?) (a? — ay +o?) = xt + ay? + of, 

(0? + y? +22 — avy —y2—2ze) (e+ y4+2=07°+ 742-8 awyz. 


IMPORTANT IDENTITIES. 81 


70. Division Identities. From the relation of division 
to multiplication, every multiplication identity gives rise te 
at least two division identities. The following division 
identities are of importance: 


1. [x°+ (a+ 6)x+a6]+(x+a)=x+6. 
2. (x°4+2ax+a’)+(x+a)=x+a. 
3. (x4°—2ax +a’) +(x—a)=x—a. 
4. (x°—a?)+(x—-a)=x+a. 
oD (xe—y)+(x-yaxt+ayty’. 
6. G+y)+ (x+y) =x x+y’. 
7 xi—yie(x—-y)= x4 xytay' ty’. 
It should be noted that in each of these identities, the 
quotient might be the divisor, and the divisor the quotient. 
(1) (#’—64)+(7%+8)=? 
This is an example of type 4. By a comparison with that 
type we see at once that the quotient is x— 8. 
(2) (a@—11 4+ 30) + (#—6)=? 
This is an example of type 1. 
a=—6; ab=30; hence, b=—5. 
The quotient is #— 5. 
(3) 27 —a*)-+(8—a)=? 
If we notice that 27 = 3°, we see that this is of the form of 


type 5. 
Hence, (27 —a®)+ (3 —a)=9+3a+a% 


EXERCISES. 


Perform the following divisions by comparison with 
type forms: 
1. (a?— 2a—63)+(a+7). 
2. [(2 x)*— b*] + (40? — 0”). 


Oo ON AA Pw 


THE ESSENTIALS OF ALGEBRA. 


(9+6a+a’)+(384a). 

(8 a + b®) + (2a+5). 

(81 a? — 25 b) + (9a+5 5B). 

(1 —10a+4 25 a7) +(1—5a). 
(y? + 11 y— 26) + (y+13). 

[(a + b)? — 64] + (a+b—4). 

[27 a — (@—y)*] +[8a—(a—y)]}, 
[(a@+ y)?—4 ab?] + (@+y+2 ab). 
(at — 14 a — 51) + (a — 17). 


- [@+y)+11@+y) —60]+(@+y=—4). 
- [et+y)’—16@+y)\a+b) +48 +b) ]+[@+y) 


—4(a+b)]. 


. (0 —9 x” — 112) + (a — 16). 
. ("+16 3” + 64) + (y" +8). 


[(a + 6)? —6 a(a+b) 4927] +(a+b—32). 


. (a — 25 yy?) + (a + 5"). 


[wt y)™— (a+ d)"] + [e+ y)"— (0 +)" I, 

[ (ax + b)? + 12527] + (av + 6)? — 5x (ax + db) + 2527]. 
[ (aa? + ba + c)? — (lx + m)*] + (aa? + bu +c+lex+n). 
(y®" — a) + (y" — 20"), : 


) (y*™ eat a) 5 (y” hal a"). 





[((V8a+y)?— a] +(V38e+y—a). 
[( Vax + 6)? — 647] + (Var +b—4y). 


: [(V aw + by)? + 125 a®] + (Vax + by +5 a’). 


CHAPTER VI. 
FACTORING. 


71. Products, Factors. Numbers which are multiplied 
together to form a product are called factors of that product. 

For example, in 5 x a x b=5atb, 5, a, 6 are factors of the product 
5ab. Only expressions free from divisions and roots will be con- 
sidered in factoring. In 7a(b+c)(x+y+2), 7 isa numerical factor 
or numerical multiplier, a is a monomial factor, b + ¢ is a binomial factor, 
and «+ y+ 2 1s a trinomial factor. 


In multiplication, we have the factors given to find the 
product; in factoring, the product is given to find the 
factors. 


72. The Degree and Number of Factors. The degree of an 
algebrate monomial is the number of letters composing tt. 

Thus, a%b is an expression of the third degree, being made up of the 
product of axaxb. a*x*y is an expression of the fifth degree in 


a, x and y; it is an expression of the second degree in a, also in z; 
it is of the first degree in y. 


The degree of an algebraic polynomial is the highest num- 
ber of letters found in any term. 
Thus, z?+ 3 2+ 4 is an expression of the second degree, containing 


xxx in its highest term. az?+bz is an expression of the third degree 
in a and z, but is an expression of the second degree in z. 


The number of factors of an algebraic expression is not 


greater than the degree of the expression. 
83 


84 


THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 


Determine the degree of the following with regard to all the 
letters : 


iT; 


73. Monomial Factors. 
may be written down by inspection. 


Thus, 


also, 


oe + an. 


2 UY +2. 

3. wtoyt+y’. 
4. 
5 


e4t3e°7+42-+ 16. 


te? + be Le. 


10. 


= eS 


e+ y+2—3 vyz. 
ao? — a’. 

v4? ata 
abe-+- 74-0 =e 
abe + b’ca + cab. 


Factors of monomial expressions 


abaz=a-a-b-a-a2; 


DMPPeo.E EY + YY. 


Here, 5 is not an algebraic factor in the sense of determin- 
ing degree; it is a numerical multiplier. 

Monomial factors contained in polynomials may be seen 
as the inverse of the distributive law. 


(a+6+c)m=am-+ bm-+ em. 


Reverse this identity, and we have the factors of 


am +bm-+em, namely m anda+6+e. 


Factor: 


Sela ee Oe ON fT 


14 a*aty. 
— 5 wy"s. 
2 xyz, 

§ a?b?c°. 

— 10 a®be. 
12 ar ye", 


EXERCISES. 


as 


4 abe! 


8. 40°4+7Tv=a2(4e+7). 
9. ax? +ay. 


10. 
11. 
12. 


3 ax + 6 aa 
5a?+10ab+5 abe. 
3 xy — 6 ay? +7 aeys 


FACTORING. 85 


13. 3ma+4 ma? + 2 may. 17. 5a®yz + 30 a’y’z — 40 ayz’. 
18. 7 a’a*yz +7 aay e +7 a®ay2’. 
19. 8a@beVae+y—6ab'oVa—y. 
20. 5 a’yzVax +b 

16. Savy — 15 aay + 21 a®ay’. +10 ay22Vax—b. 


14. —2lx+4ly+6 ly. 


15. avy + aayt+ 3 azz’. 


~ 


TTY Pits. 


A great number of algebraic expressions may be fac- 
tored by comparison with some known form of product. 
The identities of the preceding chapter are reversible ; 
when so written they become Types in factoring. 


74. The Type x’ — a’, the Difference of Two Squares. This 
expression is recognized as the product of «+a by x—a. 


Hence, 
x’ —a?=(x+a)(x—a). 


The difference of two squares equals the product of the 
sum and difference of the square roots of the two numbers. 
Thus, a#—16=(4%#+4)(w#— 4); 
also, (a+ b)?—16=(a+6+4)(a+6—4). 


Factor: EXERCISES. 
1. a’ — 40%. 6. 25 a27y? — 36 22. 
oan — 9b’. 7. (e+y)— a. 
3. a’b? — c’. 8 («+3y)— 2. 
4. 16a°b?— 250’. 9. (3x—2y)?—(e+a)’. 
5. 4—9 2”, 10. 4(x+2y)?—9(a+b)*. 


M. et—Y=CW-Y)@+Y=e@+yYVNA-yCt+y. 


86 THE ESSENTIALS OF ALGEBRA. 


In Exercise 11 we have two first-degree factors, «+ y and 
a —y, also the second-degree factor, a?+y?. No factors of 
a +4-y? can be found unless radicals be aes Such a 
factor may be called irreducible. : 


Factor : , 

12. 42° — 16 (@—34)*. 14. (w+ y)*— 4 ay’. 
13. 16—a*. 15. (8%+4y+45)—9. 
16. (— Sor oy pe (w+ y)’. 

17. (lv+ my +n)? — 4 (ax + by +c)”. 
18. (aw + by)? —4 (lz + mw)’. 
, 19. (aa)? — (b"y")?. 22. (x + 3)? — 
20. 7" yy" — 2 4p", 23. (Sa+a)’?—9 Bb. 
21. («@+1)?— a’. 24. («+a)*— (a+ 6)* 


25. (@+3y)?—(S8x+y). 


* 75, The Type x°+(a+6)x+a6. This expression is the 
product obtained by multiplying +a by «+6. Hence, 
x°+ (a+ 6)x + ab= (x +a)(x+ 5). 

Examples belonging to this type assume the form 

x’ + sx +p, | 

where p is the algebraic product of two numbers, and s 
is their algebraic sum. 

If s and p be integers, the factors of p may sometimes 
be found by inspection such that their sum shall be s. 

Thus, to factor 22 + 62+ 8, we must find two factors of 
8 whose sum is 6. These are seen to be 4 and 2. Hence, 

v+6xr4+8=(1+4)(%+ 2). 

To factor. a2?+ 10a — 24, we must find two factors of 

— 24 whose sum is 10. These are 12 and —2. Hence, 


a?+10a— 24=(a—2)(a4+12). 


FACTORING. 87 


It should be observed that if in a®+sx+p, p be post- 
tive, the factors of p chosen must be of like signs, and if s be 
positive, both factors of p must be positive. If s be negative, 
the two factors of p must be negative; if p be negative, one 
factor of p must be positive and the other negative, and the 
sign of 8 shows which is numerically the larger. 


EXERCISES. 

1. Factor 7? — 5y — 24. 

Here the factors of —24 are —1, 24; —2, 12; —3, 8; 
—4,6; and also these numbers with their signs changed. A 
pair of factors must be chosen whose sum is — 5; this is seen 
to be 3, —8. Hence, 


f= Dy — 2A'=(y 4+ 8)(y — 8). 
2. Factor (x + 2)?— 5(a + 2) — 14. 
The two factors of —14, whose sum is — 5, are —7 and 2. 
Hence, 
(w+ 2)? —5(@+2)—14= (44+ 24 2)@+2-7) 
= (x + 4) (a — d). 
3. Factor (2?+ 32)? — 8 (a? + 3a) — 20. 
The factors of —20, whose sum is — 8, are —10 and 2. 
Hence, 
(a? + 32)? — 8 (a? 4+ 32) — 20 = (a? 4 3a + 2) (24+ 3a%—10) 
={(@+1)(@+2)}{(@+5)@—2}. 


Factor: 

ara 3 ot 2. 9. v+5ay+6y". 

5. 2—3a44+2. 10. v’y’— 3ay— 10. 

6. a?+a2—2. 11. 143 a7 —102’y7, 

7. v—x— 2. 12. 62° —5ay+y’. 

8 a’o7’+ 5 az +- 6. 13. (w+ y)’+ 9(a@ + y) + 20. 


88 THE ESSENTIALS OF ALGEBRA, 


14. (a+ 3b)?— (a+ 3b) — 20. 

15. (67+ 35y)’+10(6%+43y) + 16. 
16. (+ 4")?— 6 (a+ 9’) — 27. 

17, (270+8y+4+5/4+9(2e243y+5) +18. 
18. (@+ 5x)?+10(a’+ 52) + 24. 

19, a'— 15 a+ 36. 

20. (ax + by)?+ 8 (ax + by) +7. 

21. (ax + by)?— (1+ m) (aw + by) + Im. 
22. (a+ 6a)?+17 (a+ 62a) +72. 

23. (a@’—5a+4)— (@’—5a+ 4) —2. 
24. w"™—104"+ 16. 

25. (ax)"+ (1+ m) (axv)”+ Im. 

f 76. The Types x°+2ax-+a’ and x*°—2ax+a’*. These 
two trinomials are perfect squares of 2+a and w—a, 
respectively. 

x 4+2ax+ a= (x+a)(x+a)=(x+a)’. 
x°—2ax + a°= (x —a)(x—a)= (x—a)’. 
The sum of the squares of two numbers, increased (or 


diminished) by twice the product of the numbers, equals the 
square of the sum (or difference) of the two numbers. 


EXERCISES. 
1. Factor 2? +62 + 9. 


Here x and 9 are the squares of # and 3, and 6 @ is twice 
the product of 3 and 2; hence, 


ao +6e+9= (@+3)(@+ 38) = (@ + 38). 
2. Factor 927+ 62-41. 
9e°+6e4+1=(862+4+1)(8241)=(82+4+1)2 


FACTORING. 89 


3. Factor 16 a? + 40 ab + 25 b2 

16 a? + 40 ab + 250? = (4a)?+ 2(4.4)(5 bd) + (5 bY’ 
=(4a+5b)(4a+45b)= 4a+4+ 5D)’ 

4. Factor y? — 16 yz + 64 2”. 

y —16 yz + 642 = y’? — 2(y) (82) + (82)? 

Se 8 2) — 8) aa (Yin 8 2). 

5. Factor (w? + 4a)? —4(a@’°4 4a) +4. 

In this expression we may consider xz? + 4. as a single quantity. 

(?+ 4a)? —4(a? +42) +4= (e+ 420) —2(a?+4 x) (2) + (2)? 
= (a + 4a— 2)(e’ +42 — 2) 


= (a + 4a — 2)’. 
Factor: 


6. «7 —S8x-+ 16. 
7. 40°—12274+ 9. 
8. ax? + 10 ary + 25 7”. 
9. 49 a*b?— 14ab +1. 
10. 100 — 20 ab + a?7b’. 
ll. (aw + b)? +2c(aw+ b) +’. 
12. (6e¢+4y)?’—6862+4+4y)4+9. 
13. 40? 4 99? — 12 ay. 
14. (ax + by + c)’?+ 8 (aw + by + c) + 16. 
15. (2+ y’)? —2(a? + y?)2 + 2. 
The next three exercises are squares of trinomials. See page 73, 5. 
16. 74+ yY4+24+2 ay + 2yz2+ 2 2m. 
Peo oe — 2 ab + 2 bc — Zea. 
18. 40° + 9y +22412 ay + 6y2+4 20. 


90 THE ESSENTIALS OF ALGEBRA. 


19. v7"4+12 2" + 36. 

20. yi" —14 "4.49. 

21. a” + 12 a"b™ + 36 0*. 

22 (wt+y)"—G6a(a+y)"+9 a. 


77. The Types x°?—y’ and x°+y’*. It has been shown by 
actual multiplication that 
YS (x — ye + ay ty); 
and O+ p= (x+y) —xy+y’). 


EXERCISES. 
1. Factor 8 a? — 27 b*. 
8 a? — 27 D8 = (2 a)’ — (8B)? 
= (2a—30)[(2a)’?+ (2a) (86) + 6 b)*], 
= (2a—5b)(40+4+6ab+49 0"). 
2. Factor (a+ y’)? — 8 ay’, 
(a? + 9%) — 8 ay = (a? + y?)8 — (2 ay)? 
=(@'+y')—2ay]l[@+y) + @ + yx) 2a 
+ (2 xy)”] 
=(@—y)[@ +9)? +2 ay (+ y’) + 4 xy’). 
3. Factor (x + 5)? + 8 b% 
(x+5)?+ 8b? = (+ 5)? + (2b) 
=[(+5)+2b][(w~+ 5)? — (#4 5) 264 (2b)"] 
=([%+5+426][(@+5)?—2b(@+ 5) +4 6" ]. 


Factor: 
4, 2° —8 Be, 8. («+ y)?—125 2a. 
5S 0? a0 Tobe. 9. (8x2+4+4)+8 y% 
6. a’x*® + 64. 10. (w+ 32+ 4)%— 64. 
7. ofa — 27 yf, 1l. (av + by)? — ca’. 


12. (8a%+4y)>+2a+y)% 


FACTORING. 91 


Certain expressions may be transformed into the sum or 
difference of two cubes, and the factors then found. 
13. Factor «* + 7°. 
In this case we may write 
ab + y= (0°) + (yy 
= [a*+y"] [@*)? — Gy’) + YY" 
= [2+ y?] [at — vy? + y']. 
14. Factor 2&—y*. 
As in Exercise 13, we may write 
gf — y= (a)? is (y?)3 
= (x? iy y’) f(a")? ae ha an (y”)?] 
= (#+y)(@—y) [t+ vy t+y*] Psst? 
=@+y)@-y@t+ayty’) (a? —ay +47), 4, OG Fe 
(See Exe ay f 42 page 80.) 4 





Of course, the factors of 2° — 7° could hav en obtained by 
comparing with the type 2 — a. Thus, Wy he 
fi 


eva @y— yy fe ly 
ee ty)G@—y) gh! 
=(@+y)@—a ty’) ery +2y +y%). 
Other types of binomials, such as “at — bt, a? + b°, and so on, 


may appear for factoring, but such special cases will not be 
considered at this time.!'. © 


| ” 


Factor : | ; 
15. Sa — b°. 18. a®u’+ (y+2)% 
16. (ax)® + (by)®. | 19. (ax + by)* — (cz)% 
17. 2° — 1. ; 20. 64 a°+ (bc)®. 


21. [(w + Wii yy ls 
27 ae + 3 ab? + b%)? + c%, : 


3 


92 THE ESSENTIALS OF ALGEBRA. 


78. The Type Ax°+Bxy+ Cy’ or Ax°+Bx+C. If the 
first of these expressions is capable of separation into 
factors free of radicals, it must be composed of two 
binomials of the form az+ by and lx + my, where a, 6, J, 
m are algebraic numbers, 2.e., they may be + or —, inte- 
gral or fractional. 

Multiplying these supposed factors together, we have 


(ax + by) (le + my) = alz? + (am + 61) xy + bmy?. 
Hence, to factor Az*+ Bry + Cy? we are to find four 
numbers, a, 6, 1, m, such that 
al= A, bm=(C, and am+6l/= B. 
This method is illustrated by the following examples : 
(1) Factor 62?+ 31 xy +35 y’. 
The factors of 6 are 6, 1 and 3, 2; the factors of 35 are 35, . 
1, and 7, 5. <A good plan is to arrange the letters thus, 
was WiC DD 
c+ y 
Now attach the factors of 6 and 35 to a, y, respectively, as a 
trial arrangement. Let us place them thus, 
Sa+Ty |. 
“2a+5y 
The square terms appear correctly, 6 2’, 35 y, but the cross 


products, 3x-5y and 2a“-7y, do not add so as to give 31 ay. 
Hence, our trial arrangement is not correct. Let us try 


2e+7T7y) 

ou+5y) 
This gives 6 a?+(2x54+3x7)ay+ 35 y’, the correct product. 
Hence, 62°+4 3lay4+35y=(2247y)(82+5y). 


FACTORING. 93 


The case in which y is equal to unity, giving Av’?+ Brt+ 0, 
requires no special mention when the factors may be found by 
inspection as above. 

(2) Factor 3 a+ 16 ay +5 7’. 

In this example we are to find the arrangements of the 
factors of 3 and 5 with the letters x, y in 

| a+y) 
aty)’ 
such that the sum of the cross products shall be 16 ay. 
_ By trial, the arrangement is seen to be 
eves 

| det ¥ 

Hence, 32°+16ay+5y=(w+5y\(Ba+y). 

(3) Factor 152°+ 58 2+11. 


The factors of 15 are 3, 5 and 1, 15; the factors of 11 are 1, 
11. Our trial arrangement may be | 


ye a iD 
15%+ 1 
But the cross products 11 x15+1 x1 do not give 58. Another 
| trial may be 30411 
5eo+t i |, 


which gives for the middle term 3a+550=58 a. 
Hence, 152°+58e%4+11=(82+11)(2+1). 


Factor: EXERCISES. 

1. 30? +7 ay+2y’. 4, 120° +2 vy —2 7’. 

re 6 a? —5 ay —6 7’. 5. 3 ax’ + (9+ a) bx + 3 b* 
3. 21¢’+3lay4+4y7". 6. 20 7° — 22 yz + 6 27, 


Tek —2Gu2 — 8 2, 


O4 THE ESSENTIALS OF ALGEBRA. 


79. The Type ax’?+6x-+ce. ‘This expression is called 
the General Quadratic in a single variable x. For differ- 
ent values of a, 6, e this expression represents every 
quadratic that may be written. Thus, with a= 5, 6=6, 
c= 2, we have 627+ 627+2; with a=4, 6=—5,c=—l, 
we have 427— 5a — |, etc. 

We shall illustrate the method of factoring the general 
quadratic by a few special examples. 


(1) Factor 2a? + 16a — 20. 
By dividing out 2, we get 
2x°+ 16% — 20 
= 2 (a? + 8a — 10) 
= 2[a°+8 «+16—10—16], adding and subtracting 16. 


= 2[(a + 4)? — 26], rearranging terins. 
26 may be written (/26)?, and we have 
20 +162 — 20 


= 2f (a 4+ 4)? —(V26)"), the difference of two squares. 
=2[4+4+ V 26] -[e+4—~26], the product of sum 
and difference. 


Note. Adding 16 is called Completing the Square of the first two 
terms. ‘To complete the square of x? + 2 mz we must add m?; 1.e., we 
add the square of half the coefficient of x. To complete the square of 


x? + 6 x2 we add (S)= 32. To complete the square of 2? + kx we add 


Complete the square: 
1. a’ + 52. 


Here (3)? must be added. 
If we wish the expression to remain unchanged, we must 
also subtract (3)’. 


FACTORING. 95 


Hence, e+ 5e=a+ 5a 4 25 — 25 





a) 
= (w+ $28. 
2.  — 32. Y aed Wi ete a Sr 
3. a +7 x. 8° 5a? +7 
4. vw’ — 8x. 9. 3a’— 82. 
5. 380°4+9x=3(a? +32). 10. 72? — 35a. 
6. 52? — 25a. 11. 92?— 25 a. 
(2) Factor 5a? 4 15a — 10. 
527° +152—10 
= 5[a?+ 3a — 2], dividing by 5. 
= 5[a?+3a+ i — ($)?— 2], completing the square. 
= 5[(x+ 3)? -— 2-2], rearranging. 
=5[ee+ 471 
= 5[ (a + 8)? — (V1)"], difference of two squares. 
= 5[a +54 ek E pores Sl product of sum and 
p; difference. 


(3) Factor 42° + 6a + 2. 


4o? +6242 
= 4[a?+ 3a+41), dividing out 4. 
=4[a’?+23e+ (3)?— (3)?+ 4], completing the square. 
=4[(#+3)?— 3], rearranging. 
=4[(@ + 3)’— (4)’], rearranging. 


=4[4+2+41]-[v%+2—1], product of sum and difference. 

=4[e%+1]-[v+4]. 

According to the method here illustrated we may factor 
the general quadratic. 


96 THE ESSENTIALS OF ALGEBRA. 





ax? + ba+e 
= af 2 + es + al dividing out a. 
a a 
yale b \2 6 \2 
=) 2 + —- 2+ | — ce , completing square. 
a 2a Qa 
ay b6\2 &—4ae si id 
= al (x + ) — Pe combining terms. 
= al (2 +5 ~)- “(ae 4 “) ‘| difference of two 
4 at squares. 
my Vb? — 4 ac b ia 
=2+3,+~ G2] [7 +208 


Hence, to factor any quadratic in which a, 6, ¢ have 
been replaced by numerical values, we need merely to 
replace those letters in the general factors above by the 
special values found in the given example. 


(4) Factor 3 a?+8 a+ 2. 


Here a=3, b=8,c=2; hence, we have 

















3a + 82+ 2 

8 , V64—4.3.- 8  ~V64—4-3-2 
=3 Nest: g4 2. 

Nae gone la 2am ] 

on te 

(5) Factor 5a?— 7a +4 3. 

In this case a= 5, b=— 7, c=38; hence, 
Dei eS 

Ti eo ee —7 V(—7—4-5-3 

=5 fs 

Ate Bie aio al e+55 OE 








v7 Th Gan Let ae ee 
=s]e Tiyan | ets lle 10 Ge aT i 


FACTORING. 97 


Note. The factors in this case, involving the square root of a 
negative number are called imaginary. Such numbers as v — 11, 
V —5) V— a, etc., do not belong to the algebraic number system ex- 
plained in Chapter I. 


(6) Factor — 32°+ 4a +4 2. 











Here a =—3, b= 4, c= 2; hence, 
—8attdet2=—afaty cit +2 
2-(—3) 2.(— 3) 
oe 
ealT ee) ae) 


coe a |S 24 us ry tien etl 
agh Sb 
u 


=—3 [6-2-2 Vi) [2-245 1/40}. 
jets EXERCISES. 
226 7? — 25 a — 15. 6.0ll ae — 55a +99. 
2. —22?4+142—10. 7, —427°—282+4+ 32 
3. 32°7—52+4+9. 8. 62°—3027+4 48. 
4. 7% —3524 49. 9. —82°+40 2-8. 
5. —2’+11e¢—3. 1o. 10 x?— 702-4 20. 


Il. FACTORS BY REARRANGEMENT AND GROUPING 
OF TERMS. 


The method of factoring certain algebraic expressions 
will often be suggested by a proper rearrangement or 
grouping of terms. Two general plans of grouping are 
worthy of attention. 


98 THE ESSENTIALS OF ALGEBRA. 


80. Grouping with Regard to the Descending Powers of 
Some Letter. 
(1) Factor 42° + 7 +244 2y +4224 2 yz. 
If this example is not recognized as a perfect square, we 
should proceed thus: 
4 ary +244 vyt4 w242 ye=stet4a(yt2tyt+2 yz4+2 
= 42°44 x(y+z)+(y+2)? 
= (2x2+y+2). 
(2) Factor 2? + 6ar—8 br +9 a?—240ab+16B% 
Arrange with regard to the letter 2, giving 
+6 ax—8be+9e@—24ab+16P 
H=27+2e(3a—4b)+9e0—24ab4+160 
=H=r+2e8a—46)+(8a—4b) 
=(r+3a—4by. 
(3) Ractar 4+T7axr+6 br+10 a? + 21 0b4+9 8% 
Here we may arrange with regard to the letter a. 
v4+T7axr+6b2+10a?+210ab+98P 
=10a°?+Ta(#+3b)+2°+6b2495b? 
=100°+7 a(x+3b)4+(e4+3 5)? 


=(5a+2+3b)(2a+24+35). 
“hae | EXERCISES. 
1. 2¢0+3ar+2°4+7ab+50°+4 be. 
2. 22° -8 +3 y+ bu—2 by +T ay. 
3. 7 4+47+192—4ay—12 yz+6 22. 
4. 4¢°4190?416°+4+12 ab—16ac— 24 be. 
6. 22°—37—32—2y—52z+10 yz. 


FACTORING. 99 


81. Grouping with Regard to Some Letter that Enters in 
One Degree Only. By forming a product of a number of 
factors, one of which contains a letter not found in the others, 
we shall see the purpose and application of this method. 

Let us multiply together the following: 

(a#+353a+ b)\(7#+a4+™m). 
The product is 
e+4ar+3a*+be+ab+m(r+3a+b). 

It will be noticed that m appears in but three terms, 
and that these terms when collected constitute m times the 
first factor of the product. Hence, if a literal expression 
contain a single letter entering to a single degree only, the 
coefficient of that letter contains a factor of the gwen expres- 
sion, if the expression has any factors. 

An example will illustrate this method. 

Factor 32?+8ay+3ka+5ky+5y’. 

Here k enters to a single degree. 

Arrange with regard to k, and we have 

8024+ 8ay+5y74+(8e+5y)k. 

If the expression can be factored, 3%+5y must be one of 
the factors; hence, 32?+8 xy +57? must containdx+5yasa 
factor. 

827 +4+8ay+5 743 ket bky=3v+8ay+5Y+k(8e+5y) 
=(52+5y)(a#+y)+kh(5e+5y) 
=(Se+5y)(x+y+h). 


EXERCISES. 
Factor: 1. 40?+4ay—357?+5ay—2 az. 
2. Ptezr—4yz—ay—127y%. 4 wyt4ay42?—ar—20 a7 
3. 4actbe—?412@—ab. 5. 142°47 pe—Sayt+py—y, 


100 THE ESSENTIALS OF ALGEBRA, 


82. Binomial Factors by Trial. Let us divide 27+627+5 
by x — m in the ordinary way. 
x—-m)e+6e+o0(4+6+m, quotient. 
x2 — mx 
(6+m)x+5 
(6 + m)x — m(6 + m) 
m+6m + 5, remainder. 

It will be noticed that the remainder found on dividing 
a? +62+5 by «— mis precisely the value that the divi- 
dend becomes when z has been replaced by m. Thus, if 
in 2@+6a2+5, we put z=™m™, we get m+6m+5, the 
same as the remainder above. 

This remainder is, of course, true for any value of m. 
We make a few special illustrations for various values of m. 


(1) What is the remainder on dividing 
7 x“ +6xe+5 by «—4? 


Our result above shows the remainder to be the value of 
v’+6ae+5 whenx=4; hence, the remainder after division is 


44+6x4+5=45. 
Verify by actual division. 
(2) Find the remainder after division of 
e’+6xe+5 by «—5, 
Replace aw by 5, and the remainder is ' 
°+6x5+5=60. 
Verify by division. 
(3) Find the remainder when 2? +6a-+5 is divided by a+ 5. 
Here, m=—5; hence, the remainder found by dividing by 
x+5is (= by 6 (= bya eb = 0) 


FACTORING. 101 


The remainder being zero shows that # + 5 is an exact divisor 
of 77+ 6 + 5. 
(4) Find the remainder on dividing a —8#+12 by «—8. 
The remainder is (8)? — 8(8) + 12 = 12. 
Hence, « — 8 is not an exact divisor. 
(5) Find the remainder on dividing x —8#+12 by «—6 
The remainder is (6)? — 8(6) + 12 = 0. 
Hence, x — 6 is an exact divisor, 7.e., a factor. 
In factoring by trial, the number of trials is limited to the 
number of divisors of the constant term. 
(6) To find the factors of 2° — 7 a — 6. 
The factors of 6 are +1, +2, + 3, + 6. 
When divided by x—1 the remainderis 1— 7—6=—12. 
When divided by «+1 the remainder is —1+ 7—6=0. 
When divided by x—2 the remainderis 8—14—6=—12. 
When divided by «+ 2 the remainder is —8+ 14—6=0. 
When divided by x —3 the remainder is 27 —21—6=0. 
No other divisors need be tried, for we already have the 


three (© +1), (w+ 2), and (# — 3), and an expression of the 
third degree can not have more than three factors. 


EXERCISES. 
Factor each of the following: 
Lo — 2? —42+4 4. 
Sea — 42 — 8. 
ee oe — 13 2 — 15. e+ 15 2 — a2 — 15. 
ag — 6 a + 11 « — 6. edo 47 & 60; 
5. 2 —102°4+ 312—30. 10. 2—122°+4 482 — 64 


gt 2'e* — 9'o — 18. 
a +9 a7 26 o + 24. 


oO ND 


102 


26. 


27. 


a I 


2. 


THE ESSENTIALS OF ALGEBRA, 


EXERCISES IN FACTORING. 


bu? — b. 13. 27 — 642°. 
10 em? — 40 c*. 14. 16 wy? — a. 
a? + vy + wz + yz. 15. a — 20° SG, 
lm + mr — lr — 1°, 16. m*—4 mn? + 4 nil, 
20° +3ay—2az—3yz. 17. (a+b+c)?—(a—b—c)’. 
§—5a+a’ 1s. 14a®—8a’—21a+4+12. 
at + ay? — 72 v4. 19. p?—2pq+q?—7. 
m—4m + 96. 20. 252?—10ay—92?+ y”. 
hab? 2 a7b Has 21. 9a?— 3600+ o608 
oe Aa 1: 22. a? -+-29a +120; 
. 2P—8(m+ 1) 23. ot 412 27? — oo 
. 64at — 81 4. 24. 247418743. 
25 W(e@—y)+3aly—x) + 2(e—y). 
pa te, 28... abc? — 2 a®b’c? + atd?c*. 
BL afl 29. 9x°—6ay—1+y’. 
m> + m> — m? — 1. 30. 646*-—127—1; 


REVIEW EXERCISES. 


Remove parentheses and simplify 
2e—3sy—j{5ae—[8y+5a—(4a+y—38a—4y)]}. 


Put «=5 and y=1 in the above exercise, and find the 


value. 


3. 


From the sum of 3~a—8y+2z2 and 5y—Ta—3z2 take 


their difference. 


4. 
5. 


Multiply out (a* + 4) (a? + 2) (a? — 2). 
Divide a" — Say + 3a"y— x? by a®—y. 


FACTORING. 103 


6. Factor az’ + ba? —a— be. 
7. Simplify 
(a + a)? — (@—a)’?—[(@ + a)? + (a — a)(% — a) — x” J. 
8. Divide a*+4y' by 2 —2ay42y’. 
9. Two numbers differ by 17. One third of the smaller is 


one greater than + ofthe larger. Whatarethenumbers? (Let 
x= smaller, «+ 17 = larger.) 


10. Divide (@+y*)(@—y’) by #—2a’y+ 2 ay’ — y'. 
11. Factor «*—102?+ 9. 
12. Add with respect to xv, ax +a(b—c)a—y'r+i11o. 
13. Find the value of 

w+ yz —yfa— (By +22) (5a+y—6)—2a—y}, 


waen oO, y7—2, and z= 1. 


14. Find the value of Spee when «=— i. 
a 2) — x 


15. (2a®— 32+ 5)? = what ? 
16. Divide at? — gm yrth 4. gmtt yn — yfnt) py gm 4 yn, 
17. Multiply a”*b + a7b"? by a™°b+ ab". 
1s. Multiply (a*+1)(a"—1)(a"+4+2a"+1). 
19. Verify a(a+1)(a+2)(a4+3)=(@+4+3a41)?-1. 
20. Irom 
(w@+y+2)(a+y—2) take a? —{y—[2y°—(—2ay+2)]}. 
21. Find the sum, difference, product, and quotient of 
Aa(y—z)ynr* and a(y—z)n"*, 
22. Factor a®’—4a7+a-+6. 
23. Divide 8a? —7°+2°+6ayz by y—2-—2. 
24. From 4{(@—3y) —1(9y— 22) take 7, (74 —9y). 


ivide 1a?—12a?+1q—1 Pipe se 
25. Divide ta’—tia’?+14a— by ta—}. 


CHAPTER VII. 
DIVISORS AND MULTIPLES. 


83. Highest Common Factor. 


A Common Factor of two or more numbers is a factor of 
each of them. 


a is a common factor of ax, a?y, and ab®. x—yisa 
common factor of a7 — y? and 2 — y®. 


The Highest Common Divisor of two or more numbers is 
the product of all their common factors. 


ax is the Highest Common Divisor (H.C.D.) of a®zy, 
3 a222, and 5a2rz. a* is common, and so is z. ' 

In arithmetic, the term Greatest Common Divisor is fre- 
quently used. ‘This term is not applicable in algebra. In 
the above example, a2 may or may not be greater than a. 
If a is less than 1, then a? is less than a. Hence, in alge- 
bra the term Highest Common Divisor is used. 


84. Highest Common Divisor of Monomials. Rue. Zo 
the Greatest Common Divisor of the numerical coefficients 
affix each letter common to all the monomials, and to the 
lowest power it occurs in any one of them. 

S.aratyee?. 4 aaaeiegs fy age eee 

The G. C. D. of 3, 4, and 5 is 1. 

The common letters with the proper exponents are 
a4, a3, ¥?, 2. . 

Hence, the required H.C. D. is 1 a%a3y%z. 

104 


DIVISORS AND MULT(PLES. 105 


EXERCISES. 
Find the H.C. D. of: 


A afy’2, 8 xyz”, 12 abyz’. 

5 abe’, 10 a*b®xay, 25 a7b*cx. 

16 a®aFytz, 48 atba®y*zt, 36 a®btary"Z, 

14 a®b?c*m?n?, 21 atb*m4, 42 a®b?mén’. 

22 p®qta’y, 44 p*tg?a°y’, 66 pq®a’y®. 

14 (a — b)x*y, 18 (a — b)°a*y’, 12 (a — b)Paty? 
15 (@ — y)*2*, 21 (@— y)*2*, 33 (a — y)*2". 

18 (a? — b*)ay, 27 (a? — b°)*2*y, 36 (a? — b°)?xty’. 


oN OO PF oO D FP 


85. Highest Common Divisor of Polynomials. The H.C.D. 
of polynomials may be found by factoring. If each factor 
is considered as a single quantity, the method is the same 
as used in finding the H.C. D. of monomials. 


(1) The H.C.D. of 2#’7—82+7, a?—1, 2?4+3a—4 is found 
as follows: Bs et 7 aio Ter 1), 
ae? — 1 = («+ 1)(@—1), 
a +3e—4= (x4 + 4)(e—1). 
It is seen at once that #—1 is the H.C.D. 
(2) Find the H.C. D. of 
e+5e—140, «e*—8a, 2t—42°+42, 
w+ 5x2?—144%=2(x + 7)(a — 2), 
ot — 8x = «(v’?+ 24+ 4)(x— 2), 
et—4a° 44a? = a(x — 2)(x — 2). 


Here we see that w(« — 2) is the H.C. D. 
The H.C. D. is sometimes used in reducing fractions to their 
lowest terms. 


106 


\ 
\ 
THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 


Find the H.C. D. of: 


1. 


pon anh OD 


ee ee ee 
PON HF O 


15. 


ao? —y?, ow —Qay+y?, x? — ay. 

a’? — b*?, a@—ab*, v?+2ab+0%. 

a —y’, w+’, w+ xy, 

e—Te+12, +2e2—15, 2’?—9. 
w+8a+15, @—2a—35, a®?+3a—10. 
b?—140+449, b?+b—56, b? —b— 42. 
30°—12%+412, (w4— 2)8, 3a?—12. 
a*b? — b*, ab? + 6°, ab — b?. 


a — aty!, a8 (ay — 9?) 


. (8—3 2"), x — avt— 62%. 

. 0 —2 0° — 3504, a? — 25 ot, 

+ 2ey+y, e+ y7+3ay(e+y). 
. m—3m—7T0, m?—11m?+10m. 


x? — xy + x2 — yz, ry —y’. 
a®’— 8, atb?—4a°b?, 4a?—16a+16. 


86. Lowest Common Multiple. 

When two algebraic expressions are so related that the 
first is an exact divisor of the second, the second is said 
to be a multiple of the first. 


6 abcx is a multiple of 2 abc, because 2 abe is a divisor 


of 6 a®bex. 


A Common Multiple of two or more algebrate expressions 
ws exactly divisible by each of them. 


12 a®x37323 is a Common Multiple of 8 a®zry and 4 a?yz?. 

The Lowest Common Multiple (L. C.M.) of two or more 
algebraic expressions is the expression of lowest degree which 
ws exactly divisible by each of them. 


DIVISORS AND MULTIPLES. 107 


We use the term Lowest Common Multiple in algebra because we 
are concerned about the degree and not about the numerical value. 

a®z is the L.C. M. of a? and x; a‘z is also a multiple, and if @ is less 
than 1, a*z is numerically less than a®z. 


87. Lowest Common Multiple of Monomials. RuLE. To the 
Lowest Common Multiple of the numerical coefficients affix 
every letter found among the monomials, and to the highest 
power it occurs in any of them. 


3 ax, 5a*xy, 4ay*. By the above rule we write 60 a2ry4 
at once as the L. C. M. of these expressions. 


EXERCISES. 
Find the L. C. M. of: 
LZ ax, Sa'wy, 4 abey. 3. 8 pqr’, 24 n’q’r, 12 pg*r*. 
we tao we, LO araty. 4. 10 Pm'n, 15 Pmn’®, 25 l*m?n. 


5. 7 ab®xy, 14 a®b*y*, 21 ab*ary. 

6. 3(a—)b)ay, 6 (a— b)*a*y, 12 (a — b) xy”. 

7. 4(@— y)*ab, 5 (@— y)’a°b*, 10 (x — y)?(ab)*. 

8. 40 (a®— «)*y’z, 60 (a? — x)*y2z*, 120 (a? — @)*y"2". 

88. Lowest Common Multiple of Polynomials. ‘The L.C.M. 
of polynomials may be found by factoring. Consider each 
factor as a single quantity and proceed exactly as in the 
case of monomials. 

The L.C. M. of (@ — y)*, (28 — y®), a® — 6ay+5y* is 
found as follows: 

@—yY=@-WY@-y), 
(P-YP)=@-WYeEtryt+y), 
(?—6ry+5y*)=(@—-y)(a#—5y). 
The L.C.M. is (a@—y)*Q@?+ay+y7)(@—5y). 


108 THE ESSENTIALS OF ALGEBRA. 


In general the L.C.M. should be left in its factored form ; 
that is, its factors should not be multiplied together. 


The L.C. M. is used to a limited extent in reducing fractions to a 
common denominator. 


EXERCISES. 
Find the L. C.M. of: 


1. a— 0, @—2abd4+ 0, c—ab. 8. °—5r4+6, °+5r—24. 

2. x? —16, 2?—9x2+ 20. 9. P+4a*, at—16 2%, 

3. p?— 25, p? + p — 30. 10. Y —9y4+14, yr —4 y— 21. 
iy pane Wr dues beth} 11, 8B, gt = 

5. a@—4ab?, at—2a°*d. 12. 3 — 27 d’, ? —cd—6 a’. 

6. 60, +3, &—Ssa. 13. 2—a, 4—2, 442°, 16-27 
7. m’?+m+i1, m—1. 14.3406, 9—b*, 27-8. 


15. «—3, +3, 2—627+9, +6249, 2?—9. 


CHAPTER VIII. 
2 FRACTIONS. 


89. Algebraic Fraction ; Numerator ; Denominator; Terms. 
An algebraic fraction is an indicated division. 


a+b, av + by, (a+6)+e. 
It is usual to write these indicated divisions thus: 
qa at} 
he by p) 


a. atl lw 
; is read a over 6, the fraction a over 4, or a divided 





, or a/b, ax*/by, (a+ b)/e. 


by 6. The preferred reading is a over 6. 
The dividend its called the numerator, the divisor the 
denominator, and the two together the terms of the fraction. 
In the fraction ~, z is the numerator, y the denominator, 


and a and y the terms of the fraction. 
Any expression may be put into a fractional form by 
writing it with a denominator 1; a= a py sree, 
Since a fraction is an indicated division, we know that 
- x 6 =a; for the fraction 7 may be regarded as the quo- 
tient of a+ 6; but the quotient multipled by the divisor 
_ equals the dividend. Hence, ; per Diets 
109 


110 THE ESSENTIALS OF ALGEBRA, 


90. The Sign of a Fraction. The sign of a fraction is 
placed before the line which separates the numerator and 
denominator. —* is read minus the fraction x over y- 


y 


Since a fraction is a quotient and the terms are dividend 
and divisor, the sign of a fraction is determined precisely 
as the sign of the quotient in division is determined. 

sit n n n 
ers == ras reas eaten =a a 
+d d —d d +d do — d 

A fraction preceded by a minus sign is equal to the same 
fraction preceded by a plus sign, provided eather the numera- 
tor or denominator be preceded by a negative sign. 


a a — th, 
Thus, Be Fee Nas atte os. 


If the sign of either term of a fraction be changed, the sign 
of the fraction ts changed. 


Let : be a fraction. Change the sign of n, and it be- 


comes hes a change the sign of d, and it becomes 
eee ey 
—d d 


Tf the signs of both terms of a fraction are changed, the 
sign of the fraction is unchanged. 


If the signs of m and d are both changed, the fraction 
” becomes —~ =”. 
d —d d | | 

From the above it is evident that the value of a fraction 


is unaltered by changing the signs of both terms, or by 


FRACTIONS. ph i 


changing the sign of one term, provided in the latter case 


the sign of the fraction is also changed. 
ee 2h ine et 3 
d —d’ —d- i 
The line separating numerator and denominator acts as 
a vinculum on both terms of the fraction. 
ae means (x + y)+(a+b). 








ab a = a a a 
Se aa ————- = —————- @™ 
Cc c a—2Z2 —a+2z 


L— a 


91. Law of Signs. If the terms of a fraction are made 
up of factors, the signs of an even number of factors in 
either or both terms may be changed without affecting 
the sign of the fraction. If the signs of an odd number 
of factors in either term are changed, the sign of that term 
is changed, and hence the sign of the fraction is also 


changed. a(— ©)(— y)(@) _ aryz 


Cc Cc 


(a—b)(b — e)(e — a) vs (6 —a)(b—e)(e— a) 
ey 2 )(2— 2) (@—-y)(y-2)(@-2) 
_ (b—a)(e — b)(a— @) 


— (y—a)(y—2)(@— 2) 


92. Reduction of Fractions. <A fraction is reduced when 
its form is changed without changing its value. 

Reduction of fractions depends upon the following 
principle : 

Multiplying or dividing both terms of a fraction by the 


same number does not change its value. 














LIZ THE ESSENTIALS OF ALGEBRA. 
Proof. Let an es 
€ 


Then n= Ta, by multiplying both sides by d. 
nxm=fdxm, by multiplying both sides by m. 

















: x teay, by dividing both sides by d x m. 
xm / 
Hence, a= 9 sei since both = ve 
xe 
Al n sie eae Mm 
y d d+m 
Proof. Let “=f. 
roo e age 
Then Wie FO. 
n+m=fd+m, 
NM 
d-+m ee 
Hence, 5 = 55 a 


Fractions are reduced to. higher terms by multiplying 
2, , 
both terms by the same number; ~ = a (by multiplying 
both terms by y7z). et! 


Fractions are reduced to lower terms by dividing both 
Vital foe 


= -© (by dividing both 
ahedy? ays § y dividing bo 





terms by the same number ; 
terms by a®2”). 

A fraction is in its lowest terms when its terms contain 
no common factors. 


To reduce a fraction to its lowest terms, divide or cancel 
all common factors out of its terms. 


~. 


é 


FRACTIONS. 113 


@ (a? — 6?) a(a— by’ 


by dividing out a?(a + b). 


Selecting and canceling the common factors can gen- 
erally be done mentally. 


% 


@ 


Ho 6 Ve 2) 8) i 
a A Bree) (ent 3) 


(a — a) (a — b) 


(aaa) 


a— 3 
a oes 





(a—ax)(b—am) _ 


22)(8— 2) erates 


EXERCISES. 


Cancel factors common to numerator and denominator in 
the following: 


di. 


a‘b?a : 
Te whe 
3 aPa7y 
15 a®ay? 
AT ab?a? | 
51 b*cx? 
xt — ay? 
gt — yf! 
Dien 
(a + 3) (@ — 2) 


= 20.0 








(a — 5) (w+ 5)a® 


eee 1 ; 
—627+9 

a“? — 1 

Fogg tN 

ay — xy? 

aty — wy? 





aba 





20. 


21. 


ax — bx 











ah” 
7. gp) 
a? + aa 
as oles a2 
gen 
a — 4? 
BO oad yp 
"8 13 
at yt 
(w+y)° 
(a+ y)(@—y)(@—2) . 


(vw — 2) (y— x) (—#—Y) 
w+ 3a+2 
ree 
ao —5a+6. 
a? 4+a—12 
2 at + ty — 
(w+ y)(@—y) 


114 THE ESSENTIALS OF ALGEBRA, 


po ta eens a7, © 2 ae 


(e+ y +2)" ee 
so Bi sera ae) oa w+ 60% + 11a? 4 62% 
20 + Oe (a + 1) (a+ 5a + 6) 
ay a? —(a+b)x+ ad. 29 at — 60° + 11a? — 6a 
' gt — (b4+-0)a@ + be " (w — 8) (a —38@ + 2) 


ao —Te+12 1+a+b+ab 
x” — 9a + 20 (1 — a’) (1— Bb’) 


sts a? + (a — b) a — ab “31 av —OP+0C+4+ 2ca 
| #@—(b—c)x—be ' +b? £2 abe 


25. 30. 





93. Proper and Improper Fractions. A proper fraction is 
one whose numerator is of lower degree in a named letter than 
the denominator. 

g4+2¢+5 
g+327— Ta +4 
ator is of second degree in 2, while its denominator is of 
the third degree. 


is a proper fraction because its numer- 


An improper fraction is one whose numerator is of degree 
equal to or greater than the denominator. 
e+3sa%+5xr—4 
An improper fraction may be reduced to an integral 
expression and a proper fraction by dividing the numerator 
by the denominator. 
V+2¢r+1)e8+4+3e+5r—-4e4+1 
B+2e4+e 


is an improper fraction. 


ve+4er—4 
e+ 2er4+1 
2x2—5 
SHB ey Marae ee oe 92—5 


Veghts Dae. e+ 20241 


FRACTIONS. 115 


The process is similar to the arithmetical process of 
reducing an improper fraction to a mixed number. 


EXERCISES. 


Reduce the following to integral or mixed expressions: 


1. 


94. 
Since 








74+ x C= te oe 12% 
————-e 5. —_——__---—_—____—————e 
x e—4 
e—de+tt e+ 2 avy + y? + 2? 
eT erat 6) 
aw (w+ y)" 
3a? —5x+1 : aw — y —3 vy (a —y) 
e+ 1 ; a ; 
at — oF + a+ 5. 8 4at—6 0? + 12¢%+45_ 
e+ ot x , a 1 
4 
Prove =) + a + a+ a 
—2x 1—2 


Tse? divide as in Exercise 9 to four terms. 
x 





} 1 66 6 66 6 6c 6 


1—S2 


Reduction of Fractions to a Common Denominator. 
a fraction is not changed in value by multiplying 


both of its terms by the same number, we may make the 
denominator any number we please by properly selecting 


our multiplier. ; and 2 may be made to have the com- 


mon denominator 6d. 








Cred Tad 
6 bxd bd. 
CEE S50 sear. 
d dxb bd 


116 THE ESSENTIALS OF ALGEBRA. 


The method of reducing fractions to a common denomi- 
nator is the same as in arithmetic. It may be stated as 
follows : 


RuLE. Find acommon denominator, in general the lowest 
common denominator (L.C.D.). Divide it by the denom- 
inator of each fraction, and multiply the terms of the frac- 
tion by the quotient. 

(1) Reduce 5 
the lowest common denominator. 

The L. C. D. is (6 — #)(6 + 2). 

L.C. D.+b—av=b+a, 
L. C.D: +6+2=6—2. 


to equivalent fractions having 








and, 
L b+ 





Oey a(b + 2) 
b—x (b—2x)(b+2) 
Aoet c(b —& 
ba (b—a)(b+2) 
Ax an 


(2) Reduce to equiv- 


x 
v—W (©—2)(@—3) @+2)@+3) 
alent fractions having the lowest common denominator. 
The L. C.D. is (@— 2)(% + 2) (@— 8) (@+ 8). 
L. C. D. + (a — 4) = x” — 9, 
L. C. D. + (@ — 2) (@ — 3) = (@ + 2) (« +3), 
L. C. D. + (w@ + 2) (@+ 3) = (a — 2) (@— 3). 


a(a?—9) x — Ox 
@—D@—9) @—H@+D@—3)@+43) 
4 w(x + 2) (a +3) 4 oF + 20 7? 4+ 242 


(@—2) (@—3)(@+2)(@+3) @—2)(@+2)(@—3) @+3) 


(2 — x) (@ — 2) (a@— 3) —¢+7¢7—16e+12 7 
@+2)(@+3)\@—HE—3) @=2 CHEE) 


FRACTIONS. 117 


EXERCISES. 


Reduce to equivalent fractions having lowest common 
denominators : 
































Sem cee 5 3 4. 
1 -, = ae d =» ee : 
meu od. 3b ey C x+y ah x—y 
2. 9 4 and I : 4. a pei and us 
ey “x+y e¢ wt e+ 2 
ee lan as OL tg, 
e+l «—F «+2 x—2 
6. ee aR sal , and 5D 
e—4 “£42 e— 2 
x oe 
7 ——__—_—_ ae ea tae EE 
x?7—62+8 ea e? —9 oa + 20 
8. a Nye SUMS and i a a 
e+be+6 +4243 v4+3e¢+2 
a c 
9. : 
go? — y? ot yf? a? — avy” 
2 2 
ee and | 
ey w—ay—2y" w+ sy —2y? 
ee PP and 2%. 
t—x 1+2* 1-2 b--s 
12. 2 eet a and Rise Tey 
eP4+5e+6 #+62+8 e4tT7e#+12 
1 2 3 
ee ed 
a? — xy + y" x? + ey + y" xt + ay? + y* 
a 3a 5 6 
14. —— -— 15. ——__—_—_-. dt ee 
b"c” Ortiemtl a" (a+b)? an x"? (a+b)* 


95. Addition and Subtraction of Fractions. From division 
we know that 
Sree + d—e—f a 
ee 9 Brgang 





118 THE ESSENTIALS OF ALGEBRA. 


If we read this identity from the right, we have the 
result of adding a number of fractions. Hence, we con- 
clude that a, b_a+b 





| ape ee NP 
Ruue. Zo add or subtract fractions, reduce them to the 
same denominator and then deal with the numerators accord- 
ing to the rules for addition and subtraction, writing the final 
result over the common denominator. 
1 2, 32 


a—x at+z a— 





-== what? 
A a 


The common denominator is (a — a) (a+ @). 











1 a+ 
a—x @—2 
2  2a—2¢ 
ase a Cate oe 
3D on 


Fei’ Mier haem sin 
Thus we have 


a-nu (2a— 2a 3842  a+u+2a—22—32 3a—4e 














ORs hat an Gs a” — x age 
EXERCISES. 
Combine and simplify : 
1 et ey 10 
x—1 BK 
A: an 3 6 etat+ a 
x+1 (x + a)? 
5 a? Bias 
3. — ° WON toe 
x + 10 Mate 
, 
ie as ea, a ee 1 


x+2 ota! acd jae 


FRACTIONS. 119 


32+ bare ae A 6 + 41 























j OT. he Te e+ a? e+ a’ 
1 1 a b 3 ab 
10. ; Pn i ea oe lea 
Pe 1 Be Ceo a= 6 a 
1) a Tat emer ED ea, 
ee 2 a — 2 e=-9 e¢—5a+6 
Mpoerd be? 1 4, fet Ge 8 a4 
er (1)? e—1 @+20+1 2-1 («—1) 
16 Le SO Ss ES eam eae 
@—N)@—-)e+H @-HeEH 
17. 2 2 a e+2 ; 
(e+ 3)(@—2)(@+1) (#%+4)(a’—-a — 2) 
it 1 
ee 
a) el a ci De 
1 i 


IPL) ot a ee alee 
e—5e+6 w#-—624+8 


* G—NU-) | W-)E-%)  E-9e—H) 


96. Multiplication of Fractions. 


To prove a Xess 
bd bd 

Let 5a Sy (1) 

and < = f, (2) 
Ait 2 

Then 5 x Gali Sy 

But from (1) a= 7,6, (5059 
and from (2) Gas) othe (4) 


120 THE ESSENTIALS OF ALGEBRA. 


Now multiplying Equation (8) by (4), member by . 


member, ac =f fra. 
= fifo by dividing both sides by 6d. 
But 5 x ate 
er re 
Hence, b x ie bd 


Rue. The product of two fractions is the product of the 
numerators over the product of the denominators. 


This covers all possible cases of multiplication involving 
fractions, for all integers can be put in fractional form. 


Dy ain ee te 
b pe wa) 
Ae ee 

AEC leur ven. 


Any number of fractions are multiplied together by 
placing the product of all the numerators over the product 
of all the denominators. 


In all problems in multiplication of fractions free use of 
cancellation should be made. 


sai ene 
a — Be xe—-y «uty 241 











In this, if we cancel the common factors, our work 
appears thus: 
ria * Ly ee “P41 2+1 





FRACTIONS. 121 


EXERCISES. 


Perform the operations indicated, reducing the fractions to 
lowest terms: 


32 10y a a a? 
1 Se pe er oe iin oe a See 
FO 1 : 7. (= 1)x($-+1)= 5—1. 


18 


oy 1b 

3, 3(@+y) , 6@—y)" 
2@—y) I@+ry) 
Sea ety. 


o 


© 
TAT BN EG. oN 


Sia Sia Sie SIs 
| 
Q}Se 
oe 


b@+yy @™a—y 
12aa%y 14 dy) Pty) 4 
16 bay? 6a = x(a@+y) 

Sa(e+y) 2/x—y\ (¢ y 
6. 22] erage Sharing 
6b(a@—y)? 3\e+y inset 





- 


a 
S!1S 


a 


L; 





See 

S 

x 
Re, 
QO 1a 

SS ae =, 
tS 

x 

Fg 

RAIS 

2 

is 

x 

Cees 

QQ 
Se 

S 


a 
sie 
MS” 


= 
Q 
oe 
aus 
o1S 
——— 
| 
Mo 
ols 
ee EH 
Qe 
ees oP 
+ 
Fe 
Qa) 
eae 
ecaewttadl 
a) 
+ 
Qa 


2 
14. (E-2). 15. at;tt)(5-1). 


Tepe ea: 4) x Soe Lie (Cancel common factors.) 
e+i5ie+6 (e+ 1)(@+ 4) 


w+ Te+l2 w+6a+5 | 
oft 4a+3 ~2?4+9a4 20 


uae aed me eat SD 
e—Te2+12° 2? —62+5 


e-E 1\? 5 (at— 25 e—1 
19. . 
oa Aaa ee) 
or —4oa"+4 _ ym—1 
Se as ae aa 


17 


18. 














132 THE ESSENTIALS OF ALGEBRA. 


97. Division of Fractions. 


To prove SSSR 8 
Greed) 0 Wane 
Let goty (1) 
say (2) 
Then Stat acti 
Butfrom(1) a=f,d, 


and from (2) Clea ifad- 





, ES Rie ake 
Co Wa Weta 
ra zs = by multiplying both sides by a 
els Oras b. 
ay d ae ad aes d ae 
os (bu WEbO We pee 5 
Hence, ; “ . — 7 x <. 


Rue. To divide one fraction by another, invert the divisor 
and multiply. 


Since all integers can be expressed in fractional form, 
this rule suffices for all forms of division involving the 
fraction. 





(1) i Nios he a ee 
y Ye wl yar aay 
This is dividing a fraction by an integer. 
9 .2_a4,@_a ¥ _ ay, 
2) yf gh age (ee x 


This is dividing an integer by a fraction. 


FRACTIONS. 











Byes. 7:2 ae P Rekcaek £7 ey. 
(8) Se B= x= 4-9). 


ety w—-y wt+ty a- 


123 


In this case the common factors (a — b) and (#+y) are can- 
~eeled. The student should constantly be on the alert for 
common factors and cancel them as soon as they appear. 


EXERCISES. 



































1 San, Gara ae +36 ab+30? 
Baty 10 ay? -@+56 ° a +5ab 
> Aara’ | 8 aa 6 —Ay oR cto 5 ary 
"15 b%y? ~ B3by eae 20? — 8 ay 
3, Le atylat A any? gq Sty . @ryy’ 
" “Qatyd 3 atyz — @—y? (a —¥)? 
pee et ves (2 t-Y/)" Poe alo hipy ems See 
(e—y) w-—y v—y’ a? + ay + ye 
eo .«—sy 
Ly ae a Y. 
(@+3y) a+3y 
10 (a+a)(v—b) , #+(a—b)x—ab 
 @—a)(@+b)  2’—(a—b)x—ab 
ee Die ee OC ase Oe 
e—y* (a+y’) b°—(c+a)?> —a+b—e 


98. The Complex Fraction. A fraction which has a frac- 
tional expression for either or both of its terms is called a 


complex fraction. 








a lina 1 
eas pal, ames Fe x+- 
Bo 6x y 
1,0 a+ 
ae ey Y 


are complex fractions. 
ax 


Since a fraction is an indicated division, a complex 
fraction is simplified by performing the division indicated. 


124 

















a 
URN wet. Oe, 0.400 
terse pb \ic ane 
a 
aa al | Se ie 
(2) ey ey wy wy _y—® 
sa BES ytu ay 
oy ty xy vy 
bear xy 





EXERCISES. 


Simplify the following: 

















at 5 — +4 
es 1+ - 1+2 
sey y 
" 6 
e+——5 
ee a 3 
2," eee 
oie 2a gl og: 
Kae w+a_a—a 
3. x Be A ak wa 
ies aes ens Meg yee | 
x %—-a «e+a 
4 x+y a sy 
a Op Oe 
5 RecN kart 8! 
a+y AFane he ral 











THE ESSENTIALS OF ALGEBRA. 





bose 





10. 


11. 


12. 


wy y+e y+e 





a+ 3 ey+ 27° 
a Daiet 
ety wx+2y 








CHAPTER IX. 
EQUATIONS IN ONE VARIABLE. 


99. Identity; Conditional Equation. Distinctions be- 
tween an identity and a conditional equation have already 
been made in Chapter II]. We have had illustrations 
of the identity in all the fundamental operations and in 
factoring. | 
(w+ 2)r+4=277427+4, 
and (e+1)(w@—5)=r7?—-427—-—5 


are identities; that is, they are equalities that are true for 
all values of x. tn 

A conditional equation restricts the value of some one 
letter, the letter so restricted being called the variable. 
The other numbers of an equation are called constants. 


(_) z«=5 restricts x to the value 5. 
(2) 8x=18 restricts x to the value 6. 


(8) ax=6 restricts x to the value us 
a 
(4) 2?=42 restricts 2 to the values 0 and 4. 


Each of these equations becomes an identity for the 
restricted value of the variable. 


Thus, (1) becomes 5=5 for r=5, 


(2) becomes 3 x 6=18 for x= 6, 
125 


126 THE ESSENTIALS OF ALGEBRA. 


(5) becomes a x ° = for re v 
a a 


(4) becomes 02=4>x 0 for 7=0; 
and (4)? =4 x4 for z=4. 


100. Root of an Equation. A value of the variable for 


which the equation becomes an identity, is called a root of the 
equation. 


A root of an equation is frequently called a solution. 

Any root of an equation when substituted for the vari- 
able is said to satisfy the equation; that is, an equation 
is satisfied by any value of the variable which reduces it 
to an identity. 

22+5=15 has 6 for a root. Substituting 6 for 2, 
this equation becomes 2 x 6+ 3=15, which is an identity. 
6 is the value of x which satisfies the equation 274+ 3=15. 


101. Classes of Equations. A rational equation is one in 
which the variable is free from radical signs. 


8227—Tx+11=0 isa rational equation. 


An irrational equation is one tn which the variable is 
affected by a radical sign. 


5 a? — 6Va2=12 is an irrational equation. 
An integral equation is one in which the variable appears 
only in the numerators of its terms. 
Dee Ot ie if 


3a? + 17 + — EST is an integral equation. 


A fractional equation is one in which the variable appears 
mn one or more denominators. 


EQUATIONS IN ONE VARIABLE. 127 


(el ae 
e xv—4t2r+1 


Ba+ 


is a fractional equation. 


A linear equation is one in which the variable appears to 
the first degree only. 


32+ 7=11 isa linear equation. 


A quadratic equation is one in which the highest power of 
the variable is two. 


522+ 7Tx2—16=0 is a quadratic equation. 


A cubic equation is one in which the highest power of the 
variable is three. 


x —6a*+11x¢2—6=0 isa cubic equation. 


It should be noticed that several of these terms may be 
applied to the same equation. 


322°—Ta+13=0 is a rational, integral, and quadratic 
equation. 


EXERCISES. 


Classify the following equations : 


1. av’?+b2+c=0. 5. +3 e—Te=l, 
2. ax+b=0. 6. 3Vr=11—2. 

3. ax’ + be? +cr+d=0. 7 8 5+28 4. 
4. w+3e4+7=0. e e+1 


102. Equivalent Equations. It will be noticed that the 
equations ibys Sate 
(2) 84—12=0 


have the same root, viz., x = 4. 


128 THE ESSENTIALS OF ALGEBRA. 


The equations 
(3) 527 — 102 = 0°and 
(4) 527-102 +6=6 
are satisfied when x = 0 and whenzx=2. Such equations 


as (1), (2), and (3), (4), having the same roots, are called 
equivalent equations. 


_Hquivalent equations are those having precisely the same 
roots. 


Determine which of the following equations are equiva- 
lent: 
(1) x—1=0, whose root is 1. 


(2) 38x%—6=0, whose root is 2. 

(8) 2¢%—2=0, whose root is 1. 

(4) 22?—x=0, whose roots are 0 and 1. 
(5) 2a—4=0, whose root is 2. 

(6) z27—1=0, whose roots are 1 and —1. 
(7) 832—6+4+5=5, whose root is 2. 


(8) 2—a2—4=-—4, whose roots are 0 and 1. 


(9) ar : = (0, whose root is 2. 


2 
(10) Aes : + 1=1, whose roots are 0 and 1. 
(11) 27+3=4, whose roots are 1 and —1. 
103. Solution of Linear Equations. 7 solve an equation 


is to find all of tts roots. 


The solution of an equation consists in deriving one or 
more equivalent equations, the last of which gives the value 
of the variable. 


EQUATIONS IN ONE VARIABLE. 129 


Thus, to solve 
(1) 82+4-—2=6+4 22, 


we bring 22 to the first member of the equation and 
(4 — 2) to the second member. This is done by subtract- 
ing these quantities from both members and gives us 


(2) 8a—247+4-—2-—(4—2)=64+22—24—(4-2), 
which becomes, by omitting the terms that destroy each 
other, 

(3) 59-27-56 —4 + 2. 

This process is called transposition. It is effected by 
changing the sign of a term when it is moved from one 
member of an equation to the other. 

We next unite the terms in each member of Equation (3), 
which makes 

(4) oA, 

This process is called combining terms. It is effected by 
addition. The last equation expresses the value of the 


variable 2 Hence, 4 is the root or solution of the given 
equation. 

It should be noticed that Equations (1), (2), (8), and 
(4) are equivalent equations, each having the root 4. 

If 4 is put for x in these equations, they become 


(1) 8x4+4—2=6+2 x4, 

(2) 83x4—-2x4+4-2-(4-2) 
=64+2x4-—2x4-(4- 2}, 

(3) 3x4-2x4=6-—4+2, 

(4) 4. = 4, 

Each of the above is an identity. 


130 THE ESSENTIALS OF ALGEBRA. 


As another illustration let us solve 

(1) 4¢4+2—T=227-8. 

Transposing 2 and —7 to the second member and 22 
to the first member, we have 


(2) Ag —2og=—o- [ae 

Combining, we have 

(3) 2a=—3. 

Dividing both terms of the equation by 2, we have 

(4) B= — §. 

The root or solution of Equation (1) is— 3. Equations 
(1), (2), (8), and (4) are equivalent equations, each being 
satisfied by the root — 3. 








Putting v= — 3, they become 

e 4(-D)+2-T=%A-H-8, 
or —64+2—T=—3-—8. 

@) -(--2(- = 247-8, 
Or —6§6+38=—247—8 

(3) 2(—3)=—8. 

(4) —g=—8. 

As a further illustration let us solve 

() 328 46222! 418. 


In order to get rid of the fractions in this equation, we 
multiply both members of the equation by the Lowest Com- 
mon Multiple of the denominators. Multiply both sides by 
30, the L. C. M. of 2, 5, and 3; the result is 


(2) 45¢—754+180=62—2027+4+10+4 3890. 


EQUATIONS IN ONE VARIABLE. 131 


This process is called clearing of fractions. It is always 
brought about by multiplying both members of the equa- 
tion by the L. C. M. of the denominators. 


(3) 452-6 7+ 20 x=75—180410+390, by transposing. 
- 4) 59 = 290, by combining. 
oo). Die ap by dividing. 
5 is the root or solution of (1). 
Equations (1), (2), (8), G), and (5) are equivalent 


equations. 
Show that each is satisfied by the root 5. 


104. Rule for Solving a Linear Equation. 


(1) Clear of fractions. 

(2) By transposition bring all the terms containing the 
variable to one member of the equation and all the constant 
terms to the other member. 

(3) Combine the terms of each:member of the SE by 
addition. 

(4) Divide both members of the equation by the coefficient 
of the variable. 

Steps (1) and (4) depend upon the axiom that multi- 
plying or dividing equals by equals gives equals. Steps 
(2) and (8) depend upon the axiom that increasing or 
diminishing equals by equals gives equals. 


EXERCISES. 
Solve the following equations : 
1, 82—5=19. 4. 7x—12+3=52+416—85. 


2. (x—11 — 24. §. 12—4¢74+2=13—72+10. 
3. 3a@—5=2+13. 6. 15—142—7=17—167—6. 


132 THE ESSENTIALS OF ALGEBRA. 


7. Te—ll+42—-T=32—8. 

8. 11—5e417-—38e7=18 —11 2428. 

9. 5e«—16—6e—6=115—Ta—4a2—T. 

10. 2~—224+7274+14=62-—8+4+42-4 42—5@2@ 

11. 10(@— 3) =8(e—2). (Remove the parentheses first.) 
12. 11(4ea—5)=7 (64%—5). 

13. 3(@—2)4+2(2%—3)=3 (#@—4). 
























































14. de@+1_x+12. 20. Ze—1 , 3a—b_ Oaye) 
2 S o Z 4 
5a—4 2443 . 32 oe ae £ 
[2-0 2e+5 4ey—-2 , 8% 126 
16. — . 22 7° 28— — Sie eis 
3a ee oie °— 5 
1. S20 ge owe a +1 2x—5 c_4 
b 5 8 9 7 
© o2 22 f= ee ee x—3 
1s. —+—=6-—-——-1 24. a . 
973 3 6 i 5 3 
7 Amerie arte | etl 2a. x—9 
19. —-—-=-+=-— 25. 4= 
Pe Is: 103 9 fe a 8 
26 221430288 19452, 


27. 3(@ — 2) —2(¢#—5)+2e—20=17. 
28. ax + ba=a’?+ 2ab+ G 
(a+ b)a= (a? + 2ab+5). 
_a@+2ab+0? 
a+b : 
“oa +d. 


x 


EQUATIONS IN ONE VARIABLE. le 








29. ax+.a’?= bx + 0’. 34. ce—3a=—c? — 9. 
Sl. ag-+- ba =—a®-t b* 36. Spat 
a 
32. a’a—abe+ bae=a'?+ BF ; 
H P 37. ole at Sh = 2 0. 
33. av-+ b*= aq? — b*x— abe. atb a—b 


38. (w—3)(@+5) —T = (w%+4)(x—8). 
39. (24—5)’?4+4=(#—6)(4a—3). 
40. 14—(2—2)’=5— (x#+3)(x— 2). 


EXERCISES. 


1. One half of A’s money is $35 more than B’s. They 
together have $280. How much has each ? 


SOLUTION. 


Let x = B’s money. 
xz + $35 = one half of A’s money. 
2x +70 = A’s money. 
x+22+ 70 = 280. 
+27 = 280 — 70. 
ge aa tA 
x = 70 = B’s money. 
22 + 70 = 210 = A’s money. 


VERIFICATION. One half of $210 is $105, which is $35 more 
than $70. Also, $70 + $210 = $280. 


2. A has $10 more than 3 times as much as B, and they 
together have $250. How much has each ? 


3. Find two numbers whose sum is 81, such that one may 
exceed 6 times the other by 4. 


134 THE ESSENTIALS OF ALGEBRA. 


4. Divide 114 into three parts such that the first may exceed 
the second by 15, and the third the first by 21. 


5. Divide $176 among A, B, and C, so that B may have 
$16 less than A, and $8 more than C. 


6. Divide 440 into three parts such that the second is double 
the first,increased by 10, and the third is the sum of the first 
and second. 


7. What two numbers have a sum of 861 and a difference 
Ot 21 


8. Find a number that exceeds 31 by the same amount that 
2 of the number exceeds 1. 


SOLUTION. 


Let zx = the number. 


x — 31 = the excess of the number over 31. 


; — 1= the excess of } of the number over 1. 


By the conditions of the problem, these are the same; hence, 


6x2 — 186 = 2 — 6. 
62 —2z=— 186 —6. 


5 « = 180. 
x = 36. 
VERIFICATION. 36 — 31 = 5. 
1 of 86 —1=5 


9. What number increased by + of itself and 80 is 30 more 
than double itself ? 


1o. Hight times the difference between the third and fourth 
parts of a certain number is 40 less than the number. What 
is the number ? 


EQUATIONS IN ONE VARIABLE. 135 


11. If 10 be subtracted from a number, 4 the remainder 
+ 40 is 50 less than the number. What is the number ? 


12. Find two consecutive numbers such that 4 of one plus + 
of the other is 44. 


SUGGESTION. Let x = one number, and 2 + 1 = the other number. 


13. Find two consecutive numbers such that 4 their sum is 
34 less than the larger one. 


14. Find three consecutive numbers such that 1} the first 
+ 4 the second + } the third is 88. 


15. In 10 years John will be twice as old as Henry was 
10 years ago. John is 9 years older than Henry. Find their 
ages now. 

SOLUTION. 
Let x = Henry’s age. 
x + 9= John’s age. 
zx +9 +10 = John’s age 10 years hence. 
x — 10 = Henry’s age 10 years ago. 
2+9+10=2(2—10). 
24+9+10=22 — 20. 
%z—2xr=—9-—10-— 20. 


—xr=—— 39. 
30, 
zr+9= 48. 


16. A man’s age plus that of his wife’s is 95 years; 40 years 
ago he was twice as old as she was then. What are their ages 
now ? 

17. Eight years ago a father was 9 times as old as his son 
was at that time; in 37 years the father will be 13 times as old 
as the son is at that time. What are their ages now? 


is. A man left + his estate to his son, + to a nephew, t toa 
niece, and the remainder, amounting to $2600, to his wife. 
What was the value of his estate ? 


136 THE ESSENTIALS OF ALGEBRA. 


19. A house is sold for $2280. This is a gain of 14%. 
What did the house cost ? 


SOLUTION. 


Let x = the cost of the house. 
ii = gain. 
x + pls x = 2280. 
1002 + 147 = 228000. 
1142 = 228000. 
e200; 


20. A horse sold at a loss of 7% brought $111.60. What 
did the horse cost ? 


21. A man invests } his capital at 4% and the remainder 
at 5%. His income is $2800. What is his capital ? 

22. What number must be added to each of the terms of the 
fraction +4 to make it 24? 


23. What number must be subtracted from both terms of 
the fraction 42 to make it 2? ; 


24. Divide $5600 into two parts such that the income from 
one part at 3% may be equal to the income of the other part 
at 4%. 

25. Divide $760 among A, B, C, and D so that A and B 
together shall receive $150, A and C together $190, and A and 
D together, $ 580, 


26. $7.20 is changed into 36 coins. Each coin is either a 
dime or a quarter. How many of each are there ? 


27. A bill of $10.20 is paid in an equal number of dimes, 
quarters, and half dollars. How many of each are used ? 


28. A man bought sheep at $4 a head, calves at $9, and 
cows at $35. He bought twice as many calves as cows, and 


twice as many sheep as calves. The cost of all the stock was 
$690. How many head of each did he buy ? 


EQUATIONS IN ONE VARIABLE. 137 


29, Find three consecutive numbers such that the sum of 
the quotient of the first divided by 10, the second by 11, and 
the third by 61, is 25, 

30. Find three numbers such that the second is a times the 
first, the third } times the second, and their sum ec. 


31. One half of A’s money is equal to B’s, and five eighths 
of B’s is equal to C’s; together they have $1450. How much 
has each ? 


32. A man walks out at the rate of 4 miles an hour, and rides 
back at the rate of 10 miles an hour. How far can he go out if 
he must make the round trip in 7 hours ? 


33. A man sold 12 acres more than 1 of his farm, and had 
2 acres less than 2 of it left. How many aeres had he? 


34. A train leaves a station at 8 a.m. and rvuzs 30 miles an 
hour. At 11 a.m. another train leaves in the same direction 
running 45 miles an hour. When and where will it overtake 
the first train ? 

35. A and B are two towns 120 miles apart. A messenger 
starts from A to B at 7 a.m. and travels 10 miles an hour. At 
8 a.m. another messenger starts from B to A and travels 
12 miles an hour. When and where will they meet ? 

36. A man in traveling from New York to Buffalo, goes + as 
far by boat as by train and 1, as far by carriage as by boat. 
If the distance to Buffalo from New York be 490 miles, how 
far does he travel in each conveyance ? 


105. The Linear Type. Every linear equation in a single 
variable may be reduced to the type form 


ax+6=0. 


In this form a and 8 represent any positive or negative 
numbers whatever. 


138 THE ESSENTIALS OF ALGEBRA. 


For example, S77 Ce eee 
Clearing of fractions, 
127-284 10%—-2=827-28+432. 
Transposing all terms to the first member, 
12%7+1027-—82-—3827—28 —24+258=0; 





Collecting, IaH 20. 
This is in the type form. 
Comparing it with ax +b=0, we see that a= 11 and )=—2. 


The solution of the type form azr+b=0 


‘ b 
1S r=. 


a 
Hence, the solution of the above example is x = ?;- 
Special roots of axw+b=0. 
If 6=0, then the solution of az + 6=0 becomes 
perme ef 
a 
If a=0 and 6 is not 0, then the solution of ar+6=0 


pecomes 
r= — 


We have here a new form whose value we must in- 
vestigate. jb 


CER, 

HP = +104, 
+= +1008, 
+ © = + 10008, 


+b 
+ ___ 4 1900000008. 
0000001 mY 


EQUATIONS IN ONE VARIABLE, 139 


It appears that, as we decrease the denominator, the 
value of the fraction increases. When the denominator 
of the fraction is very small, the value of the fraction is 
very large. When the denominator becomes 0, the value 
of the fraction is large beyond measure. We express this 
fact by saying that the value of the fraction is infinity. 
The symbol for infinity is oo. 


b 
f= —-—-= — O, 


0 
Any number divided by 0 is equal to oo. 
3 i ae: 10008 _ 


oO, -=0o, ——_ = ©, ete. 


ch area 0 


If a= 0 and 6= 0, the solution of ax + 6=0 becomes 


e=—- 


0 
Lae i 0 
5-38 the symbol of indeterminateness. g may have 
any value. 


a — 
(1) 


part. 





If in the above we put x = 1, it becomes 





1-1 0 
———=1+41, or ~=2. 
Reena 0 
“ae 
(2) cae lib ase 
x—d 
If in this we put x= 5, it becomes 
25 — 25 





0 
(aS 5 5, ty a 10. 
Rb aeY Nhe ai, 


140 THE ESSENTIALS OF ALGEBRA. 


106. Equations of Second or Higher Degree which depend 
upon the Linear Type. 


If we have the equation 
x—d2+6=0, 
we may by factoring write it in the form 
(x —3)\@—2)=0. 


We know that if one factor of a product is 0, the 
product is 0. The product (# — 3)(a@ — 2) may be 0 by 
either factor being 0. Ifz«—38=0, then the product is 0, 
or if r—2=0, then the product is 0; that is, the product 
is 0 if z= 38, org = 2. 


This is called equating the factors to 0. 


A root is a value of the variable which satisfies the 
equation. Hence, in the above equation, 3 is a root 
because it satisfies the equation. 2 is likewise a root 
because it also satisfies the equation. Therefore, the equa- 
tion z2— 52+ 6=0 has the two roots x= 3 and z= 2. 

An equation of higher degree than the first may be 
solved by the linear type, provided, after all the terms 
have been brought to one member, it may be factored into 
linear factors. Each factor equated to 0 will give one root. 

Hence, the number of roots is equal to the degree of the 
equation. 


EXERCISES. 
1. vw —5a—24=0. 
By factoring, this is written (# — 8)(a# + 3)=0. 
Hence, ¢ + 8.=.0 oriaees, 
also “e+-3=0orz=—3. 


The two roots are 8 and — 3. 


EQUATIONS IN ONE VARIABLE. 141 


2. vw —3x2—40=—0. 13. 1627— 25 =0. 

3. v+624+8=0. 14. «7 —T2+10—0. 
4. ¢4+1074+16=0. 15. 2 +3e2—10=—0. 
5) a —d5x2—14=—0. 16. 2° +82+15=0. 
6. # —162+448=0, 17. 4¢7—122%49=0. 
7. @+4=42. 18. 2 —6'?+2ax+a’?—0. 
8. 2? —1=—0. 19. 0? —12 2+ 35=—0. 
9. vw —25=—0. 20. #’?— 2142+ 20=0. 
10. 2 +11 = 36. 21. 2 + 282+ 75=0. 
11. «?—16=0. 22. o —T.a= 98. 

12. 2’—(a—b)’?=0. 23. wv +7T«x—98=0. 


24. 32°+11¢—4=—0. 

25. 12(¢+1)—3(@—1)4+2?-1=0. 

26. (x — x)? — 22(a? x) + 40=0. 

27. (@’ +3 2)?— 8(#’? +32) —20=0. 

28. +5e7+62=0. 
29. o°® —12 07+ 27 x=—0. 35.92 =o —2 ab) = 0. 
30. (7+4)+(2 x—5)?=73. 36. 2-— 1327+ 36=0. 
31. (2%—5)’—(24410)?=24. 37. (a@+4)?—-@ae—1ty=0. 
32. (6%+4)—(B8a—8)’=0. 38 (@?+227)?—(@’—42)’?=0. 
33. 7 —~=0. 39. 404-8 a —52’°=0. 
34. aa’? —(b+c)?=0. 40. 2 —382+52—1=0. 

107. Fractional Equations. Certain fractional equations 


may be reduced to the linear form or to the form dis- 
cussed in the last section. 


142 THE ESSENTIALS OF ALGEBRA. 


Fractional equations are made integral by clearing of 
Fractions. 

The common multiple used in clearing of fractions will 
contain the variable. It may give an integral equation 
which is not equivalent to the given fractional equation. 


(1) nae 


x 


Clearing of fractions by multiplying by 2, we have 


2 — 1 = 0, 
or (#¢—-1)@4+1)=0. 
W hence, zx=1 and z=—1. 


These roots both satisfy the given equation. 
When z= 1, 1 becomes 1-F= 1—1=0. 
x 


When z=—1, path becomes -1-+ =-141=0. 
x 


— 


If in clearing of fractions we multiply by 2%, the 
resulting equation is 
x—axr= 0, 
or x(a —1)(x+1)=0. 
Whence, z=0,2=1, andz=—l. 


We now have three roots, two of which, 1 and —1, 
satisfy the given equation, while the other one, 0, does not 
satisfy it; for, when r=0, x seh becomes gi which 
: x 0 
is not equal to 0. 

The root 0 which is here introduced by clearing of 
fractions is called an extraneous root. 


The root 0 occurs because we multiplied by a multiple 
higher than the L.C. M. | 


EQUATIONS IN ONE VARIABLE. 143 


In integral equations any multiple whatever of the denom- 
enators may be used in clearing of fractions, but in frae- 
tional equations the L. C. M. should always be used. | 





(2) 54 ot a4, 


§a7?—54+474+1=42'?'—4, 
§at—4e2?4+7-—-54144=0. 











v+a= 0. 
a= 0Q0andz=—1. 
When z=0, 
5+ oti a4 becomes 5+ ot a4, or 5—1=4. 
When z=—1, 
Hee etl =4 becomes 5 + att a, or 5 +o=4. 
a= —1 does not satisfy the equation and is therefore an 


extraneous root. 

The root —1 occurs in this solution because the fraction 
a+1 
g2—1 


it to its lowest term the equation becomes 


1 
Dye me es 4 
AVERT aT 





was not reduced to its lowest terms. By reducing 


fe —) +1 ss 47 — 4. 
be 4e—5 2 1-4 — 0, 
me 
No extraneous root now appears. 


Before beginning the solution of a fractional equation, all 
fractions should be reduced to their lowest terms. The safe 


144 THE ESSENTIALS OF ALGEBRA, 


plan in all fractional equations is to test every root, retain 
only those roots that satisfy the equations, and reject all others 
as extraneous. 











EXERCISES. 
ee ey 3 2) ae 
ot DS wo 
SL 3 2 
2. 4 —=— —= 0. eee SS 
ta oo il ean 
Zz 1 1 
5 aa ii aan 
peer g SU IGG) a 


6. The quotient of a number divided by 7,increased by the 
quotient of 63 divided by the number,is 6. What is the number? 

7. A number is increased by 82 and the sum divided by the 
number; the quotient is ;/; of 1 more than the number. What 
is the number ? 


8. The sum of the squares of two consecutive numbers is 85. 
What are the numbers ? 
SOLUTION. 
Let x = one of the numbers. 
xz +1 =the other. 
2 (el es, 
z24724+227+1 = 85. 
2277+227+1- 8-0. 
227+227—84=0. 


g+a—42=0, by dividing by 2. 
(x +7)(x — 6) =0. 
De arm] OTT ae 


The numbers are 6 and 7 or —7 and — 6. 


9. The sum of the squares of two consecutive numbers is 
41. What are the numbers ? 
10. ‘Two numbers differ by 5, and their squares differ by 105. 
What are the numbers ? 
11. Three times the product of two consecutive numbers 
lacks 92 of being twice the sum of their squares. What are 
the numbers ? 





EQUATIONS IN ONE VARIABLE. 1465 


12. The area of a square field is doubled by increasing its 
length 12 rods and its width 5 rods. What is the length of 


one side of the field ? 
SOLUTION. 


Let x = one side of the field. 
a? met deel) (tO). 
222= 474+ 1724 60. 
242 —2?-17x7-—60=0. 
u2—17x—60=0. 
(x — 20)(4 4+ 3) = 0. 
x= 20 and x =— 5. 

Both of these roots satisfy the equation, but only 20 can be used 
in this problem, as it would not be possible to have a field one side of 
which is — 3 rods in length. 

13. The denominator of a fraction 1s 3 more than its numera- 


tor. If7is added to each of its terms, the value of the fraction 
is increased by ;3;. Find the fraction. 


14. A can do a piece of work in 10 days, B in 8 days, and C 
in 6 days. In how many days can they all do it working 


together ? 
SOLUTION. 


Let 


= the time required. 


lI 


part done in one day. 


Slee; 


II 


part done in one day by A. 


II 


part done in one day by B. 
= part done in one day by C. 


= part done in one day by A, B, and C. 


oe 
CO | 
4. 

II 


ole ole 


on Ret 
we pp Ole 


12741 


Spats 
8S 8 & Ole QIK olHKolH 


II 
oO 


146 THE ESSENTIALS OF ALGEBRA. 


15. A cistern has two pipes; one will fill it in 8 hours, and 
the other in 12 hours. If both are open, how long will the 
cistern be in filling ? cE 


16. A cistern has three pipes; one will fill it in 12 hours, one 
in 10 hours, and the other will empty it in 15 hours. If all 
three are open, how long will the cistern be in filling ? 


17. A number added to 22 times its reciprocal makes 13. 
Find the number. 


: is called the Reciprocal of x. 
z 


1s. A can do a piece of work in a days, B ean do it in Dd 
days. In how many days can they together do the work? 


19. The area of a square field is doubled by increasing its 
length a rods and its width 6 rods. What is the length of one 
side of the field ? 


20. A fraction whose numerator is 3 less than its denom-: 
inator added to its reciprocal gives 2;%. Find the fraction. 


CHAPTER X. 
LINEAR EQUATIONS IN TWO VARIABLES. 


108. Roots of a Linear Equation in Two Variables. The » 
type form of the linear equation in a and y is 


ax+by+ec=0. 


a, 6b, and e may be any positive or negative numbers 
whatever. 

The equation 22+5y—10=0 is a special form of 
ax + by+e=0; in which a= 2, b=5, and e= — 10. 

If in 2~7+5y—10=0, we transpose 5 y—10 and di- 
vide by 2, we have 
pel) 0 Y, 

2 


x 


The value of z depends upon y. It has one, and only 
one, value for each value of y. 


ia Uta 0. lfiy=—1, ¢= 4+. 
yo, «x= 3. y= —2, c= 10. 
=22=0. y= —3, 2 = 28, 
y=3, c= — 3. yi 4 tee 1D: 
at = — 5, y= — 9, w= 3B. 
y= 9, r= — 4p. Yie= 10,92 = 20. 


147 


148 THE ESSENTIALS OF ALGEBRA. 


The equation 272+5y—10=0 is satisfied by r=5 
and y = 0, for these values reduce the equation to 


2x5+5x0—-10=0. 
a= 3 and y=1 also satisfy the equation, for they reduce 
Ey xbox oe 


Therefore, z= 5, y= 0 and «= 3, y =1 are roots of the 
equation 27a +o5dy—10= 0. 

In the set of values of 2 and y above, every value of a 
and the corresponding value of y constitute a root. The 
number of such sets of values that are possible is evi- 
dently unlimited. Besides the roots (5, 0), (3, 1), (0, 2), 
Cr Bs 3), C= Dy 4), (Gatce D), (Jes val 1), ClO 2), and 
GCG), — 3), any number more could be worked out at 
pleasure. A root of an equation in two variables may 
be written (m, n); m is the value of z, and. n is the 
corresponding value of y, the two together constituting 
a root. 


109. Graph of the Linear Equation. ‘The codrdinate azes, 
or lines of reference, are two hnes perpendicular to each 
other. 

The azis of abscissas, or x-line, is the horizontal line 

WOK. 

The axis of ordinates, or y-line, 

is the vertical line Y’OY. 
Abscissas, or x-distances, are al- 
X ways measured parallel to the z-line. 

They are positive when measured to 

the right of the y-line and negative 

Y’ when measured to the left of it. 
Wia.-1: Ordinates, or y-distances, are always 








LINEAR EQUATIONS IN TWO VARIABLES. 149 


measured parallel to the y-line. They are positive when 
measured above the z-line and negative when measured 
below it. 

The coérdinates of a point are its x and y distances. 
The w distance is the abscissa, and the y distance the 
ordinate. ‘The codrdinates cf a point completely deter- 
mine it with respect to the lines of reference. 

A point is designated by its codrdinates written (m, 7). 
This means that m is the abscissa and m the ordinate of 
the point. 

Y 


P, (2,3) 





P, (—5,—4) 


Y 4 


Higa: 


The point P, or (2, 3) is located by measuring from O 
to M, a distance of 2, and from M to P,, parallel to OY, 
a distance of 8. The point P, or (—4, 2) is found by 
measuring from O to N,a distance of 4, and then from 
N tose: parallel to OY, a distance of 2. The abscissa in 
this case is measured to the left of OY because it is nega- 


150 THE ESSENTIALS OF ALGEBRA, 


tive. The location of the points P; or (—5 —4) and P, 
or (6, — 2) is also shown on Figure 2. 


BEC ns See: 
Ch 
SeEEPEE ai. 

a a 





Hd 
Y 
~ 


Suen ane 











EXERCISES. 

In Figure 3 the side of each small square is a unit or one. 

1. Write the codrdinates of each lettered point. Thus, 
A is (2, 2), & is (—6, —1). 

2. Locate on Figure 3 the following points: (5, 1), (— 2, 1), 
(3, satay (es 3, oe v (6, 1), i= 1, aoe Cass 2); (0, 3), E 3, 0), 
(0, 0), (1d, 8), (23, 22), (— 3h 4 

On Figure 4 are located the points (10, — 2), (74, — 1), 
(6, 0), (24, 1), ©, 2), C— 24, 8), C— 5, 4), (— 7, 5). 


LINEAR EQUATIONS IN TWO VARIABLES. 151 


It will be seen that these points are in a straight line. 
The points located on Figure 4 are some of the roots of the 
equation 22+ 5y—10=0, worked out in Section 108. 


oe ee Ha a 
ate 


Estee: 
a 
fs 
P| 
im 
ia 
He 





TAL__| 








of 


me MuaG 
Sie 














ay 
’ 





The graph of an equation ts the line upon which are found 
all the points indicated by its roots. 

The line MN is the graph of the equation 

The coordinates of every point upon this line satisfy 
the equation 27+ 5y—10=0. The point P is (24, 1). 
(24, 1) is a root ‘of 2x+-5y—10=0, for when x=24 and 
y =1, the equation becomes 2 X 24+ 5X 1— 10=0, which 
is an identity. 

The graph ofx—2y=4. Herezr=4+42y. 

@=—=4 when y=—0. a= 2 when y=—1. 

ee Oowhen oy = 1. z= 0 when y=—2. 

teow oen. i = 2. z=—2 when y=—83. 


Loy THE ESSENTIALS OF ALGEBRA. 


The points represented by the roots above worked out 


are (4, 0), (6, 1), (8, 2), @, —1), (0, —2), and (—2, —3) 


and are shown on Figure 5. 


y 
FEEEE EEE eae 
































Ber... 
Plo ea 
3 HERRERA 
[eh ise, 





ESS 
=e 
ae 


Fia. 5. 
The points tetera in Figure 5 lie upon the straight line 
PQ, which is the graph of —2y=4. 
The graph ofx+y+5=0. Herex=—(y+5). 
e=—)d ~whene y= ae x=0 when y=— 6). 
oe — 6 when aye: z=1 when y=—6: 
x=—2 when y=—83. 


_ The points represented by the above roots are (—5, 0), 
(—6, 1), (—2, —3), (0, —5), and (1, —6). 

Locating these points upon Figure 6, we find that they 
all lie upon the straight line AS, which is the graph 
of x+y+5=0. This line passes through the points 


LINEAR EQUATIONS IN TWO VARIABLES. 153 


Ac—4, —1), B(—8, —2), and C(—1, —4). Verify that 


these are roots of the equation. 


aaa 


R, 


——— 


oo 


(- 


Spat 
ae 
Bava 


bisaeules 
EP 
Alb 





























bad 
Eames 
te 
a 
sel 
| 


Pita at 


EXERCISES. 
Draw the graphs of the following equations: 
1 3e—2y=6. 3. 24—5y=10. 5. e+y=4, 
2. da—y=8. 4. ¢—y=4. 6. 24—3y=0. 
The graph of every linear equation in two variables is a 
straight line. 


Since two points are sufficient to locate a straight line, 
we need but two roots of an equation to draw its graph. 


154 THE ESSENTIALS OF ALGEBRA, 


The graph of 844+ 5y=15. 
Since 2=0 when y=8, one root is (0, 3)s 


and since x=0d when y= 0, another root is (5, 0). 





att 


Fia. 7. 


Locate these two points upon Figure 7, and draw through 
them the straight line AB. ‘This line is the graph of 
Sa+5y= 15. 

In general, the most convenient pair of roots for deter- 
mining the graph of an equation is found by making #=0 © 
and solving for y, and then by making y= 0 and solving 
for z. ‘These two roots give the points in which the line 
cuts the codrdinate axes. 


EXERCISES. 


By the above method make the graphs of the following 
equations : 


1. 3x—2y=6. 4. Tx—y=T. 
2. 4x—y=8. 5. 2 Tiere 
3. 2y—5x=10. 6. 357+4y=12. 


LINEAR EQUATIONS IN TWO VARIABLES. 


110. Graphs of Two Linear Equations upon the Same - 


Diagram. 
Graphs of x—y=6 and 2a+y=9. 

















PEN ae 


“A 











ea] Se 25 Ne 








PAGS 





ys : D 
Fig. 8. 


| 


Pes | lee | Ne 


155 


SS 


The graphs of these lines are AB and CD, respectively. 
They intersect at the point P (5, —1). This point P hes 
on both lines, and its codrdinates constitute a root of each 
equation. By putting x=5 and y= —1, x—y=6 becomes 
5—(—1)=6, and 22+y=9 becomes 2x5—1=9. This 


verifies that (5, — 1) is a root of each equation. 


Since two straight lines can intersect in but one point, 
a pair of linear equations can have but one common root. 


156 THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 


By means of their graphs find the common root of each of 
the following pairs of equations: 


7 \Fescie, lat , {eA ae 
laty=6d. et+y=—B5. 

a ees a aan 24+4y=6, 
x—2y=1. laty=1. 

a raat te a (e—y=1, 
e—2y=A4. (2e 4.2 ene 


7. Graphs of e+y=1 and 2x%+2y=9. 


Fig. 9. 


The graphs of these equations are shown in Figure 9. They 
are parallel lines, and so do not intersect. The two equations 
have no common, finite root. It should be noticed that the 
coefficients of x and y in the second equation are just double 
the coefficients of # and y in the first equation. 


{ x—dsy=6, 3 xe+2y=6, 
[3a—9y=9. , pea y = 2. 


CHAPTER XI. 
SIMULTANEOUS EQUATIONS. 


111. Definitions. Hqguations in two or more variables, 
having the same solutions, are called equivalent when any 
one of the equations may be changed to the exact form of the 
others. 


Thus, r+y=d 
and 3824+3y=15 


are equivalent equations; each has a root (1, 4), and the 
first may be changed to the second by multiplying both 
members by 3. 


Equations not equivalent, but having the same solutions are 
called simultaneous equations. 


Thus, c+y=6 
and. s2+y=12 


are simultaneous, having the common solution r=3, y=3. 
These two equations are not equivalent, since the first 
can not be changed to the form of the second. 

Two equations in two variables form a set of s¢multane- 
ous equations; three equations in three variables form a 
set of simultaneous equations, etc. 


112. Elimination. To solve a set of simultaneous equa- 
tions, we must so operate upon and combine the given 
157 


158 THE ESSENTIALS OF ALGEBRA. 


equations as to produce a single equation containing a 
single variable. ‘The processes of obtaining such a single 
equation are called elimination. , 

In the operations of elimination the following principles 
are to be carefully noted: . | 


(1) For any expression in an equation an identical 
expression may be substituted. 

(2) When both members of an integral equation are 
multiplied by an integral expression containing the vari- 
able, the resulting equation is not equivalent to the 
original equation. 

For example: x+d3y=4. 

Multiplying both sides by # — y, we have 

(w@—y)@+3y)=4@—y), 
an equation which not only has all the roots of a+5y=4, 
but also all the roots of a—y=0. 

(3) All the axioms heretofore given can be used in the 
processes of elimination, with the single exception noted 
above. 


113. Elimination by Substitution. This method will be 
understood by noting the following solution: 


(1) a+2y=5, 

te 52—2y=1. 
By transposing the 2 y in Equation (1), we have 
(3) e=d0—2y. 
Substituting this value of z in Equation (2), 
(4) 6(5 —2y)—-2y=1. 


SIMULTANEOUS EQUATIONS. 159 


Removing the parentheses, 


(5) 25—10y—2y=1. 
(6) —10y—2y=1-—25, by transposing. 
(7) —12y=-— 24, by collecting. 
(8) gy aut by dividing. 
Substituting this value of y in (8), 
(9) = 5 —2(2). 
(10) hoe 


The root is (1, 2). 
EXERCISES. 


Solve the following equations by the method of substitution : 





1 | a+ y=4, 4 v+4y=10, 
' (8a—4y=5. | ~(38e— y=12. 
Beets, 11, AEE OE ms seh 
22+ y=9.  (8a4+2y=T7 
Soy — 2, ne rye 
| w«+3y=114. ise homers 
2 
lx 9» 
6a+ y= 20. Outen 
5a+6y=16, 1) Bp A 
—8e+38y=— 18. eet egy=li 


© exh 


Construct graphs for Exercises 4, 5, 6, and 7 


160 THE ESSENTIALS OF ALGEBRA. 


114. Elimination by Comparison. ‘I‘he following problem ~ 
will illustrate the method : | 





(2) 4vu— y=5. 
By transposing 8 y and dividing by 5, we get from (1) 
19—3 
(3) ta ee 


By transposing — y and dividing by 4, we get from (2) 


5 

(4) ime te 

Equating the two values of 2 given by (8) and (4), 

19—3 5 

(5) a ee, 
(6) 76 —12 y= 25+ 5y, by clearing of fractions. 
(7) —12y—S5y=25— 76, by transposing. 
(8) —1l7y=-—4dl, by collecting. 
(9) y= 3. 

Substituting this value of y in (3), 

(10) pa PU eXt a2, 


The root is (2, 3). 
EXERCISES. 


Solve by the method of comparison : 


a (ie eae f4et+ y = 3, 
(2Qa+ y=3. "(| 2Qe4+8y=4. 
- eee pa tay F (3a—4y=4, 


“x—dsy=—A4, |2a+5y=10. 


SIMULTANEOUS EQUATIONS. 161 


: lai (—4¢e—Zy=1, 
(8e2+7y=21. ~(— w+5y=6. 
aoe” i a 
lat2y=b. l5a+ Piezo 
ama 4 ores Wie 
L38a+2y=12. (dsat4y=4m. 


Construct graphs for Exercises 3, 5, 7, and 9. 


115. Elimination by Addition or Subtraction. ‘he two 
problems here solved will illustrate the method. 


ei jeer +o Y= 0, 


Bical 
eo. es Ta Oy =. 


We first make the coefficients of the y’s have the same 
absolute value. This is done by multiplying Equation (1) 
by 2 and Equation (2) by 3, thus giving us 


(8) 4x7+6y= 10, 
(4) 2l2—6y= 165. 
Adding Equations (3) and (4), 

(5) 25 2 = 25. 

(6) goerai li 
Substituting this value of a in (1), 
gi) 2XL+ oY = 0. 

(8) 38y=5—-—2=3, 
(9) y=. 


The root is (1, 1). 


162 THE ESSENTIALS OF ALGEBRA. 
(1) 24+Ty=38. 
(2) 84+4y=81. 


We can make the coefficients of the a’s alike by multiply- 
ing Equation (1) by 3 and Equation (2) by 2, thus giving 


2. Solve 


(3) 624+ 21ly=114, 

(4) 62+ 8 y= 62. 
Subtracting Equation (4) from (8), 

(5) LS a ae 

(6) y= 4. 

Substituting this value of y in Equation (1), 
Gh) 24+7-4=388. 

(8) 2x2= 38 — 28 = 10. 

(9) area lta 


The root is (5, 4). 

The method of elimination generally used is that of 
addition or subtraction. ‘The method by comparison is 
merely a disguised form of eliminating by subtraction. 
The particular method to be used must be determined by 
a careful inspection of the problem. 


116. Some Illustrative Examples. 


( Dey 
i a i — 4, 
ib) 413 


Here there is no need of clearing of fractions. By addi- 
tion, y is eliminated, and 


SIMULTANEOUS EQUATIONS. 163 


a2 


3 —= 6. 
om ; 
(4) 3x2= 24, 
(5) oa; 
Substituting 8 for z in (1), we have 
gy 
oy —+“=4, 
(6) ats 
(7) pat 2=2. 
(8) y = 0. 
The root is (8, 4). 
Bee 17 
1) - a? 
. @) x ss y 24 
iy Gagan 
mony 24. 
In problems of this form never clear of fractions. 
(3) pee Ae Bh by multiplying (2) by 2. 
Gey 24 aa 
ee aye 
(4) anor by adding (1) and (3). 
(5) sr pkcgeaad Wop tyre 
(6) emia fe 
Substituting 8 for 2 in (1), 
pera LT 
1) g ty bt 
rus all YE ear ee ME 
©) y 248 8 
@) y= 6. 


The root is (8, 6). 


164 THE ESSENTIALS OF ALGEBRA. 
(lez — 37 = 10: 


In this form, in which one equation has the variable with 
a coefficient 1, use the method of substitution. 

From (1), 

(3) x=3y+10. 


Substituting the value of x in (2), 
(4) 88 y410)+5y=2. 
(5) 9y+380+5y=2. 


(6) 14 y= — 28, transposing and combining. 
(8) x= —64+10=4, from (3). 


The root is (4, — 2). 





G) oe 
- BY y 
2 UI Tene 
(2) a a 


In examples of this form it is best to clear of fractions. 


(8) 14+2274+3y=3z2, by clearing (1) of fractions. 


(4) —x+8y=-—1, by transposing and collecting. 
(5) xet+y=5, by clearing (2) of fractions. 
(6) 4y=4, by adding (4) and (5). 
(T) y=1. 

GS) ake x=4, by substituting 1 for y in G). 


The root is (4, 1). 


SIMULTANEOUS EQUATIONS. 165 


EXERCISES. 


Solve the following simultaneous equations: 





eas y =, 2.3 TH 
: ae ala? 
(2e%—5y=12, [~+-=22. 
: x 
2 { de +8y=13, a4 <6, 
eee 8 — 71. 11 ‘ 
A (fe—ty=1, o%—-—-=3 
(ga+tozy=4 
[e+7=a 
( aes , ue 
4x —_=6, 12. Y 
5 7 
: kx —— =) 
Beye 2. 35 | 
fnan 
A an ay : 735° : 
ie) 18 eo 
Bie b;. Hb y 
ax + by == 4m besa eased 
2 p. a 1 ee Tt 
' n+y ™m 
F mie? 15 (3ae—Ty=0), 
ee 7 | lge+hy=7. 
é My ae (ax — by=0, 
ug me+ ny =q. 
9. Shak 2 . f mar + ny = me’, 
Dany. bnew + my =n. 


166 THE ESSENTIALS OF ALGEBRA. 








CEE ee oe v—a , y—b 
1s. - —= —Ty— =(, 19. — “~—1[= e 
8 at 4 3e—T Yy—di es ; < ; 

syt+Ta+1 2y—32+8_5 
re aH 
5 3 
20. 
Sy—Tex+10_ sy+24+6_9 
3 5) ; 


Construct graphs for Exercises 2 and 6. 


EXERCISES. 


1. The sum of two numbers is 32, and one number is 3 
times the other. What are the numbers ? 


SOLUTION. 


Let x and y be the numbers. 


Then (1)! eee, 
and (2) xz=3y, by the conditions of the problem. 
(3) 8y+y=32. 
(4) 4 y= 32. 
(5) y =8. 
(6) ey 


2. Eight apples and 5 oranges cost 31 cents, and 5 apples 
and 10 oranges cost 40 cents. What is the cost of 1 apple and 
of 1 orange ? 


3. Three bushels of wheat cost 15 cents more than 5 
bushels of corn, and 2 bushels of wheat and 1 bushel of corn 
together cost $2.05. What is the price per bushel of each ? 


4. A fraction is equal to 3. If both of its terms are 
increased by 12, the value is then 4. Find the fraction. 


(Let x = numerator, y = denominator, ~ = fraction.) 


5. Find a fraction such that if 1 is added to the numerator 
it becomes 4, and if 5 is added to the denominator it becomes }. 


SIMULTANEOUS EQUATIONS. 167 


6. The sum of two numbers is 75. The larger contains the 
smaller 5 times, with a remainder of 3. Find the number. 


7. There are two numbers; 3 times the first is 8 more than 
the second, and their difference is 42. Find them. 


8. A man spent $225 for sheep at $3.50 a head and calves 
at $10 a head. He bought 42 head inall. How many of each 
did he buy ? 


9. Ten years ago a father was 5 times as old as his son. 
Twenty years hence he will be twice as old. What are the 
present ages of each ? 


10. A said to B, “Give me $60, and I shall have twice as 
much as you.” B said to A, “Give me $90, and I shall have 
as much as you.” How much had each ? 


11. Find two numbers such that + the first and + the second 
is 36, and } the first and + the second is 18.. 


12. There are two numbers such that if each is increased by 
5, the sums are in the ratio 5 and 11, and if each number be 
decreased by 15, the remainders are in the ratio 1 and 7. Find 
the numbers. 


13. A farmer has two horses and an $18 saddle. If the 
saddle is put on the cheaper horse, the horse and saddle are 
worth 2 of the better horse. The better horse and saddle lack 
$12 of being worth twice as much as the cheaper horse. What 
is the value of each horse ? 


14. If the greater of two sums be multiplied by 5 and the 
lesser by 7, the sum of the products is 140. If the greater be 
divided by 7 and the lesser by 5, the difference of the quotients 
is 0. Find the numbers. 


15. There are two numbers which differ by 11. One sixth 
of the larger is 1 more than + of the smaller. Find the 
numbers. 


o 


168 THE ESSENTIALS OF ALGEBRA. 


16. A number consists of two digits whose sum is 18; if 27 
be added to the number, the order of the digits is changed. 


Find the number. 
SOLUTION. 


Let x = units’ digit. 
y = tens’ digit. 
10 y + x = the number. 


Then (1) Zit iz ls, 
and (2) 10y+2+27=102+~y, by conditions of problem. 
(3) 9y—-I9x=— 27. 
(4) y—-r=—3. 
(5) 24 = 10; by adding (1) and (4). 
(6) y = 5. 
(7) alias from (1). 


107 + x = 58, the number. 


17. If to a certain number of two digits the tens’ digit be 
added, the sum is 80. If the units’ digit be subtracted, the 
remainder is 70. Find the number. 


18. A number is composed of two digits whose sum is 13. 
If their order is inverted, the new number is 4 less than double 
the original number. Find the number. 


19. A sum of money was divided equally among a certain 
number of people. If there had been 3 persons more, the share 
of each would have been $2 less; but if there had been 2 per- 
sons fewer, the share of each would have been $2 more. How 
many persons were there, and what was the share of each? 


20. A lost 3 of his money and then borrowed 4 of B’s money, 
when he had $12. At first A had 2 as much as B. Find 
how much each had at first. 


21. The sum of a number of two digits and the number 
formed by reversing the order of the digits is 110. The differ- 
ence of the digits is 8. Find the number. 


SIMULTANEOUS EQUATIONS. 169 


22. A man has a certain number of silver dollars and quar- 
ters. He notices that if his dollars were quarters and his 
quarters dollars he would have $22.50 more than he now has. 
He also notices that if his dollars were dimes and his quarters 
half dollars, he would have $1 more than he now has. How 
much money has he ? 

23. In a certain school 4 of the number of boys is equal to 
4 of the number of girls; twice the whole number of pupils in 
the school is 100 more than 3 times the number of girls. 
How many pupils in the school ? ; 

24. A and B are 45 miles apart. If they walk in the same 
direction, A overtakes B in 45 hours. If they walk toward 
each other, they meet in 5 hours. Find their rates of walking. 


25. If the first of two numbers be divided by 12 and the 
second by 15, the sum of the quotients is 12; if the first be 
divided by 4 and the second by 3, the difference of the 
quotients is 12. What are the numbers ? 


26. Find two numbers such that the sum of their reciprocals 
is 25, and the difference of their reciprocals is ;35. 

27. If the base of a rectangle be increased by 6 feet and 
the altitude by 4 feet, the area is increased by 216 square feet. 
If the base be decreased by 4 feet, and the altitude increased 
by 4 feet, the rectangle becomes a square. Find the base and 


altitude of the rectangle. 

28. If B loans A $500, A will then have 3 times as much 
money as B has left; but if A loans B $ 200, B will have twice 
as much money as A has left. How much money has each ? 


29. A sum of money on interest amounted to $ 824 in 9 months 
and to $840 in 15 months. Find the principal and the rate. 


30. If the greater of two numbers be divided by the less, 
the quotient is 1, with a remainder of 8; if 4 times the less be 
divided by the greater, the quotient is 2 with a remainder of 
22. What are the numbers ? 


170 THE ESSENTIALS OF ALGEBRA. 


31. Sixty workmen, consisting of men and boys, did a piece 
of work in 5 days and received for it $430. The men were 
paid $1.75 a day, and the boys 80 cents a day. How many 
men and how many boys were there ? 

32. Find a fraction such that when the numerator is trebled 
and the denominator decreased by 4 the value becomes 3, and 
when the denominator is trebled and the numerator increased 
by 2 the value becomes 4. 

33. In 10 years I will be 5 times as old as my son was 5 
years ago, and 2 years ago I was twice as old as my son will 
be 4 years hence. Find my age and that of my son. 

34. The lengths of two ropes are as 4: 5, and when 20 feet 
is cut from each rope the remainders are as 3:4. Find the 
lengths of the ropes. 

35. A river flows 3 miles an hour; a boat going down the 
river passes a certain point in 12 seconds and in going up it 
takes 18 seconds. Find the speed of the boat in still water 
and the length of the boat. 


117. System of Linear Equations with Three or More 
Variables. We have seen that in order to solve linear 
equations with two variables, we must have a set of two 
independent equations; in like manner, when we have 
three variables, we must have a set of three independent 
equations ; when four variables, we must have a set of 
four equations, etc. 

The method of solving a problem in three variables 
will be understood by noting the following solutions: 


(1) e-2y+2=1, 
ai (2) 3a+y—z2=4, 
(3) 2a+ty+2=12. 
By looking at this problem we see that the z’s can be 
easily eliminated. 


SIMULTANEOUS EQUATIONS. WG 


(4) 4x —y=5, by adding (1) and (2). 

(5) o2+2y= 16, by adding (2) and (8). 

(6) 8x—2y= 10, by multiplying (4) by 2. 

(7) bo == 20, by adding () and (6). 

(8) Nee ee 

(9) - y=, by substituting 2 for x in (4). 
(10) ae, 


by substituting 2 for x, 3 for y in (1). 
‘The root is (2, 8, 5). 
(1) 4a—3y4+22=8, 
ay bes 64+3y4+32=T, 
(8) 2x—6y+52=4. 
By adding (1) and (2), we eliminate y, and have 


1072+ 52=10, 
or (4) Phe en ae ie ay 

Multiplying (1) by 2 and subtracting (3), we have 
(5) G27 2, 

Adding (4) and (5), 
(6) pO ta4. 
(7) t=. 

Substituting x= 4 in (4), 

(8) | Li a2. 
(9) z=. 
Substituting ¢=4 and z= 1 in (1), 

(10) 2—3y+4+2=3. 
(11) —3y=-1. 
(12) Y= 


The root is G, 4, 1). 


172 


(Do not clear of fractions in Exercise 


( 





= eee er ee ee 
J 


RliR SiR Sik 


THE ESSENTIALS OF ALGEBRA. 


“x +2=11, 

y+z=6, 
2ae+y = 25. 
22+32= 54, 
Ty+5z=106, 
de+ 5y= 76. 


10. 


EXERCISES. 





2 83 
£ ¥ B 
«| 3 
20. aye 
ey OZ 
nt nn" 
mM n 
Phy Zz 
7 
Yo 
Gar : 
5 q 
OP gaa . 
8. po 2 ee 
ey) Daa 
ae 
9, 2 i+2=5, 
tf See 
Bo 3 
D-) 


e+ytz+w=10, 
2e—y+32+w=13, 


“e+ 3y—2z2—-—w=—3, 
(y —2y4+32—-2w=—2. 


SIMULTANEOUS EQUATIONS. 173 


11. Three men have together $750; 4 of A’s and 4 of C’s 
is equal to 4 of B’s; twice A’s is $150 more than both B’s 
and C’s. How much money has each? 

12. The sum of three numbers taken two and two are 68, 
94, and 62, respectively. Find the numbers. 


13. There are three numbers such that the first with $ the 
second is 68; 4 the first with 2 the third is 73; and the second 
with + the third is 90. Find the numbers. 


14. A number consists of three digits whose sum is 17. The 
hundreds’ digit is 3 times the tens’ digit. If the order of the 
digits be reversed, the number is increased by 297. Find the 
number. 


15. A and B can doa piece of work in 6 days, B and C in 
74 days, and C and A in 10 days. In how many days can each 
do the work separately ? 


16. A cistern has three pipes A, B, and C. If A and B fill 
while C empties, the cistern will be filled in 60 minutes. If A 
and C fill while B empties, the cistern will be filled in 24 minutes. 
If B and C fill while A empties, the cistern will be filled in 120 
minutes. In what time could each pipe fill it alone ? 


17. There are three numbers whose sum is 113; 4 the sec- 
ond is 2 more than } the third; 2 of the first lacks 3 of being 
4 the second. Find the numbers. 


18. Separate the number 180 into three parts, such that the 
second divided by the first equals 2, the third divided by the 
second equals 3, and the first divided by the third equals 4. 


19. A, B, and C have certain sums of money. If A gives 
B $100, they will have the same amount; if C gives A $200, 
he will have as much left as A and B together then have; if 
B’s money were doubled and <A’s increased by $100, they 
would then have together as much as C. What sum has each? 


CHAPTER XII. 
EVOLUTION. 


118. Definitions. Square Root. One of the two equal 
factors of a number is called its square root. 

25 = 5 x. 5, hence 5 is a square root of 25. 

Cube Root. One of the three equal factors of a number 
is called its cube root. 

64=4x 4 x 4, hence 4 is a cube root of 64. 

nth Root. One of the nm equal factors of a number is 
called its nth Root. 

a“=axaxaxa-+- to n factors, hence a is an nth 
root ola”: 

Evolution. ‘The process of finding roots is called evolu- 
tion. It is the inverse of ¢nvolution. 

From the definition we see that the square root of at is. 


4 
$ — a, the cube root of a” is a =a, and the nth root 


a 


Of a” is he 

The Radical Sign. When the sign -+/ is placed before a 
number, a root is to be extracted. The number written over 
the radical sign is called the index, and denotes what root 
is to be extracted. Thus, V16 = 4, V125 =5, V16=2, 
V32 = 2, Va" =a; the indices 2, 3, 4, 5, n, denote, respec- 
tively, the 2d, 5d, 4th, 5th, and nth roots of the number 
before which the radical sign (,/) is placed. The square 
root sign (/) is usually written -/. 

174 


EVOLUTION. 175 


The Radicand. The number whose root is to be ex- 
tracted is called the radicand. Thus, in V25, V125, 25 
and 125 are radicands. 

The root of an expression of two or more terms is 
denoted by the radical sign in connection with the vinculum 
or parenthesis. 

Vat+2ab+0 and /(a?+2ab + 62) each denote the 
square root of a? + 2 ab + 0%. 


119. Even and Odd Roots. An even root has an even 
index, an odd root an odd index. 

Ae V5, are even roots. V8, V32, Vals, are odd 
roots. 

An even number of equal positive or negative factors 
multiplied together gives a positive product. Hence, only 
positive numbers can have real even roots. An even root 
of a negative number is called an imaginary number. All 
other numbers are real. All numbers are either real or 
amaginary. 

5 Wt, V1, a, 6, are real numbers. V—1, V—5, 


Ai a 6 r : ° 
V—16, V— a’, are imaginary numbers. 





120. The Law of Signs. (1) An even root of a positive 
number is either + or —. 
V81=+9, for (+9)x(+9)=81 
and (—9)x(—9)= 81. 
V a2 i td 
(2) A negative number has no real even roots. 
(3) An odd root of a negative number is —. 


peep - 8, for (—3)(—3)(— 8) = 27. Wa B2= ~—2: 








176 THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 


Extract the indicated roots: 

















1. V25@0=+5a. 11. 625 a™b*c2, 
2. Valle = + @bVe. 12. 1000 a®b%, 
3 V16 abil 13. V400 a®)¥!8, 
4. V/169 xyz, 14. bec 
(w+ y)’ 
SAO Ty 15. Ve 343 DPC, 
6. V—8 abi = — 2 abc’. 16 «| SL 
7. Va—4ab+4b'=+ (a—2b). Vag a®btc® 
8. Va?+6ax+9 x 17. V16 a®p%e8, 
Ay Aap 18. \/— 32 aay, 
ae . 19. v/aPb%c®, 
0° New 20, /— bE, 


SQUARE ROOT. 


121. The Square Root of a Polynomial. The square root 
of a polynomial is found by the reversal of the method 
used in eae a polynomial. 

(A+ B)*= A?+2 AB+ Bis the general type form of 
the square of - sum of any two quantities, and is the. 
type form used in the reversal process. 

By comparing any perfect square, whose root consists of 
two terms, with this type, its root may be easily determined. 


get 10 9 + 25. 


This may be written 27+ 22-5 +4 52. 
It is at once seen that 4d=z and B=5. Hence, the 
square root of z74+10%+25isv+5 


EVOLUTION. BEG 


The same method may often be applied to polynomials. 
whose roots have three terms. 
e+y?+9+2ay+6u+6y. 
Arrange this in type form, according to a. 
P+ 2a(yt+3)+y+6y+9. 
This may be further arranged 
w+ 2a(yt+3)\t+(y +3), 
and the square root is at once seen to be 7+ y +3. 


EXERCISES. 


Find the square root of the following by comparing with 
the type form: 
1. 2 +1624 64. 
2. 9a°+24x%+4+16. (Write (8 2)?+ 2(8 2)4416.) 
a 2 — 19%. 81. 5. 1627+ 56 vy + 49 7’. 
4. xv —10 ay + 25 y’. 6. a’a’+ 2 abuy + by’, 
7. P42 eyt+yY +242 2042 yz. 
8. w4+9yV4t2+6ay4+22e4+ 6 yz. 
5. a—G6bc+ 0? +9 ce —2 ab+6ca. 
lo. 42°4+7°+92—4 ay —6 yz4+12 2. 
MW. a?+0?4+ 4+’? 4+ 2ab4+ 2 bc4+ 2ca+2ad+2bd+2 cd. 
12. 2 —6ay+9y+16+8 a¢— 24 y. 


122. Formal Extraction of Square Root. When the root 
can not be easily determined by inspection, we reverse 


the type form. A2+2AB+ B?| A+B 
A2 


2A+B|2AB+ B 
2AB+ B 


178 THE ESSENTIALS OF ALGEBRA. 


The first term of the root A is the square root of A?. 
The second term, B, is contained in 2 AB, and is found | 
from it by dividing by 2A. 2A is called the trial divisor. 
(2 A+ B) is the complete divisor. When this is multi- 
plied by B, the result is 2 AB+ B%, which is the part of 
the square remaining after A? is subtracted. 


(1) Find the square root of 36 #?— 144 xy 4 144 7’. 


36 a — 144 ay + 144 9? |6a—12y 
36 2 


122—12y| — ldtay+ 1447 
— 144 ay + 1447? 


Here the square root of 3627 is 6x. 122 is the trial 
divisor, which divided into — 144 2y gives —12y as the © 
next term of the root. The complete divisor is 127-12 y, 
which multiplied by — 12y gives — 144 2y + 144 7°, the 
remaining part of the square after 362? is subtracted. 
62—12y is the required square root. ‘This method is 
easily extended, as the following example will show : 


(2) Find the square root of 24+ 62?+a?—247+16. 
a+ 6 a8 4+22—247+16|a?+30—4 


a! 
Q2432/623+ 2 
§ a +9 a2 
—8 7?7—24 7416 
—8 2?— 2947416 






Oia +6 7 —4 


Note that the polynomial is arranged according to 
powers of 2. 

The square root of 2* is z. The trial divisor is 2 2%, 
which divided into 62° gives 3a, the next term of the 


EVOLUTION. 179 


root. The complete divisor is 22?+32. When this is 
multiplied by 32 it gives 623+927, which leaves, when 
subtracted from the remaining part of the square, 
—82?—247416. (274372) is now regarded as the first 
part of the root, giving a trial divisor of 222+ 6a. This 
gives —4 as the next term of the root. ‘The complete 
divisor is 22?+6a—4. When this is multiplied by — 4 
it gives —8 a? — 244+ 16, which is the part of the square 
remaining. The root is 27+ 3a —4. 


123. Rule for Extracting the Square Root: 


(1) Arrange the terms with respect to the powers of some 
letter. 

(2) Extract the square root of the first term, place its root 
as the first term of the root sought, and take its square from 
the given polynomial. 

(3) Double the root already found for a trial divisor, 
divide the first term of the remainder by the trial divisor, 
placing the quotient as the neat term of the root, and also 
annexing it to the trial divisor to form the complete divisor. 

(4) Multiply the complete divisor by the last term of the 
root, and take the product from the first remainder. 

(5) Continue this process until the other terms of the root 


are found. 
EXERCISES. 


Extract the square root of the following: 
1. af 14 974+ 42° + 20 2 + 25. 
—423 +100? —1224+9. 
Aat + 25 a? +12 a + 24a + 16. 
mee alee oF 1 AD ot 1 Oe 38 a 21 a. 
» 4at +h oy? + y' + 4 aby + 2 wy’. 


a - oO N 


180 THE ESSENTIALS OF ALGEBRA. 


9 xt 4 37 ay? + 4 yt — 30 ay — 20 anf. 
1—2e—6y4+7+6ay+9 yy’. 

Aa? — 12 2y +16024+9 924162 —24 ye. 
4b +4¢7— 123-694 oe 

ht 

7 


o ON 


y' 2 by? 4 b°y? 3 
10. = — + —_+ b 
Fait sa as Oe 


124. Inexact Square Roots. ‘The following example will 
sufficiently show the method. 
Extract the square root of 1+3 2 to four terms. 


1+8e¢|l+3e—-feP+ite 


1 
2+3713824 
382+ 32 


24+3x2—32" 





bole 


co Be 
qv 
— a2? 22934 81 at 


24+3u—3 27+ 22 a3 | At 








EXERCISES. 


Extract the square root of the following to four terms: 


oe bee § 1+597- 37 
2.1+42+827 | 6. 1— Ae oe 
3) er, 7. 1-387 +5ee 2 
4, 144427, 8. 1t+u+e?+a?+ or 


125. Square Root of Arithmetical Number. The following 
principles may be easily verified by the student: 


EVOLUTION. 181 


(1) The square of a number of one digit consists of one or 
two digits. — 

(2) The square of a number of two digits consists of three 
or four digits. 

(3) The square of a number of three digits consists of five 
or six digits; and so on. 


From these principles we can at once tell how many 
digits in the square root of a given number. The square 
root of 390625 will consist of three digits. 

The number of digits in the root may be indicated by 
separating the given number into groups of two figures 
each, beginning at the right. The left group may contain 
either one or two digits. 8906 25 and 1 98 21. 

Now solve by using the same method as in algebraic 
problems. 


Extract the square root of 390625. 


89 06 25 | 600 + 2045 


36 00 00 
1200, trial divisor 
20 
1220, complete divisor 
1240, trial divisor 
5 
1245, complete divisor 





It is usual to put the work in the following shorter form : 
39 06 25 | 625 
36 
122 | 3 06 


1245 





182 THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 
Find: 1. 104976. 2. \/ 278784. 3. VbT121, 
4. 61504. 5. 235225. 6. 1092025. 


126. Roots of Decimals. Group both ways from the deci- 
mal point, and solve exactly as in whole numbers. 

36 74.87 63 64. ‘This shows the root to be made of two 
whole number places and three decimal places. 


EXERCISES. 
Find: 
1 W18'3184, 3. /.133225. 5. +/1.110916. 
2. 8.6436. 4. >/.00300304. 6. VE: 


If the result of Exercise 6 is desired to three decimals, we may 
write it- V2.000000, and then proceed as in the above exercises. 


7. ~/5 to three decimals. 8. \/11 to three decimals. 





127. Cube Root of Polynomials. ‘The type form is 
(A + B)® = A? 4+ 3 A2B + 3 AB? + B. 

All polynomials in this form may have their cube roots 
written by inspection. a%+62%y+122y?+8y? may be 
written, 2+ 3a%(2y)+3a2 y)?+C2y). 

The cube root is readily seen to be 7+ 2y. 

By exactly reversing the type form we can extract the 
cube root of any polynomial which is a perfect cube. 

A4+3A°B+3AB+ B|A+B 
AB 


3A24+3AB+4 B?|3 A2B+3 AB? + B 
3 A2B +3 AB2+ Be 
The term B of the root is found in 3 A?B and is obtained 


by dividing 3 A?B by 3 A?. This shows 3 A? as the trial 
divisor, and 3 4243 AB+ B? as the complete divisor. 





1838 


EVOLUTION. 


Extract cube root of 


27 #® — 108 2° + 198 at — 208 28 + 18207-4824 8. 





IOSTAIP 939[du109 ‘¢-+2 FZ—-t QO+ 
ef TL—# LO =et(* F—e GE te F—e BE 
S+e BF—@ TEL +e FEL st FG | —_ TOSTAT [UMD ge Qh+ ee GLH LE =e Fe? GE 
eC FQ — FEL to QOL — | AOSTAIp oqopdutod {we OT + et 9G —¥% LG 10 
(et A) + @ FD Et 1] 
IOSTATP [LMI] 5% 15 =eGr GE 


SL BP—et GEL tet FFL —¥ FS 





el 806 — 3% SGI Hc% SOT — 
gt LG 
S+X QF— wv TET Heh 80G — 3 BEL TeX SOT — 9% LZ 





"NOILATOG 


184 THE ESSENTIALS OF ALGEBRA. 


128. Rule for Extraction of Cube Root. 


(1) Arrange with respect to some letter. 

(2) Hetract the cube root of the first term for the first 
term of the root, and subtract the cube frem the polynomial. 

(3) Use three times the square of the root found for a 
trial divisor, and by dividing the remainder by this divisor 
get the second term of the root. | 

(4) Add to trial divisor three times the product of the first 
part of the root and the part of the root last found and the 
square of the root last found. 

(5) Subtract the product of the complete divisor and the 
part of the root last found from the remainder of the 
polynomial. 

(6) Repeat this process until the root has been completely 
determined. 

EXERCISES. 


Extract the cube root of: 

et hart 12¢45. 

ee Ba 6 ot 7 of 6 ee 

of — 3 ay + 6 at? — 7 oP + 6 x7 — Say 
Ve+3¢0(6+1)4+3a(041)?+60430?4+3641. 
a@+3a%+60?+ 12004 300?+0' + 1240-12026 
27 a — 64 y® — 108 a?y + 144 ay’. 

e° — 9 o' +. 42 ot = 117 oF + 210 a? — 22 eee 

8 a®b* — 36 a°d® + 78 att — 99 a®b® + 78 a7b? == 36 aa 


OA) Poe WN 


129. Extraction of Cube Root of Arithmetical Numbers. The 
following principles may be verified as in square root: 


(1) The cube of a number of one digit consists of one, two, 
or three digits. 


EVOLUTION. 185 


(2) The cube of a number of two digits consists of four, 
jive, or six digits. 

(3) The cube of a number of three digits consists of seven, 
eight, or nine digits ; and so on. 


From these principles we can at once tell how many 
digits in the cube root of a given number. 

If the given number is separated into groups of three 
figures each, each group will correspond to a digit of the 
root. | 

For example: 95 256 152 263. There are four digits in 
the cube root of this number. ‘The group at the left may 
contain one, two, or three digits. 3365 791 and 871 625. 

An example will illustrate the method of solution and 
show that the same plan is followed as in the extraction 
of the cube root of polynomials. 


Extract the cube root of 262144. | 


2+ 3 ab+ 3 ab? + 6 = 262 144| 60+4=6 
ae = 603 = 216 000 
(8 a2 + 3ab4+0)b= 46 144 


ego 5 x 607 = 10800 
+3ab=3x60x4= 720 
+ §2=— 42 = 16 
(3 ab + 3.ab + 67)6 = 11536 x 4= 46 144 
The above shows the similarity to the general type. 
In practice the solution should appear as follows: 


262 144 | 64 
63 = 216 
Trial divisor 6? x 300 =10800| 46 144 
Cite eB = -. 720 
42 — 16 
11536 | 46 144 












Complete divisor 





186 THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 
Extract the cube root of each of the following numbers: 
1. 4913. 3. 753571. 5. 2628072. 
2. 300763. 4. 614125. 6. 1.728. 


In Exercise 6, group both ways from the decimal point. If neces- 
sary, annex ciphers to fill the last group. If the root is not exact, by 
annexing ciphers the result may be carried to any desired number of 
decimal places. 


VAD UREN stone Bat 9. .0081 to three decimals. 


8. 3 to three decimals. 10. 2.05 to three decimals. 


CEAPTER XIII. 
THEORY OF INDICES. 


130. The Index Laws. ‘The following laws for integral 
exponents have already been proved: 


rie a” x a” =— qztn. ) 


2. a” +a" =a”, when m is greater than a. 
8. (a”)” — qi, 
4. (a x ce a” ~x 6”. 
ee eo 
3 b a bn 
_ m 
6. Va" =a", when m is divisible by a. 


The laws of algebra should be general. Let us assume 
that the above index laws hold true for all values of m 
and ”, and find consistent meanings for certain special 
forms, viz., a, a~”, ip 

Any number affected by an exponent is called the base with 
respect to that exponent. 

In a”, ais the base. In (@+y)”, 2+ y is the base. 


131. The Form a®. The second index law is 
a™ au a” —_ Fy hha 
Let m =n, and this becomes 
a® + a" = a, 
But aa" = 1. 
Hence, a =a" +a"=1. 
187 


188 THE ESSENTIALS OF ALGEBRA. 


Any quantity with an exponent zero is equal to unity. 
10? 





2—1, for 2+2=1; 10> Se 
102 
ee Ct Pp (e+5)=1 


132. The Form a—". The first index law states that 
q”™ x a” — qt, 
Let us assume that this holds for all values of m and n. 
Let m= —n, and we have 


OOK = 0 ee 
ae ey 
Hence, a~" =—, by division. 
a 


a~” means the reciprocal of a”. 


1 if ie 
o°=—-—; 10 1=—- 7? = —___—_; 
3 10) ed Oe ee 
ag t= £ ; ax *y™ = —; 8 a7* = —- 
xv xy" qa” 


A factor may be removed from the numerator to the denom- 
nator, or vice versa, if at the same time the sign of its exponent 
be changed. 





B2p-By2 —52z2y2 «2 x Gte2e- 1-2 __ de 
Biker Nn igh 9 | Su ng: ty 2p a 
p 
133. The Form a’. We know that (a”)"=a™. 
PR Phi arn 
Put m = and n=q. Then (a) =(a)!=a¢=a" 


P 
Now extract the gth root of both sides of (a%)?’=a?, and 


Dp Pp 
we have az=~V/a?; that is, ay means the qth root of the pth 
power of a. 
The numerator of a fractional exponent means the power 
of the base, and the denominator the root to be extracted. 


THEORY OF INDICES. 189 


To extract the qth root of a”, divide the exponent p by q. 


p p ie 
The gth root of a? is az, or a7 =Va?. 
at xat=aitt=al, 


axa= (a2)? 


nh. 
COM eam he 
of: ed 
a=Va. 
Bee eh at We. 


We find that by assuming the generality of the endex 
laws we have consistent meanings for zero, negative, and 
fractional exponents. We will hereafter use the six index 
laws of page 187 for all values of m and n. 








EXERCISES. 
0) ec ee 
ae Oa 
Bas - mt. git ge ae gece a Op Se | z 
2. —jui 4 12T2-T—He 
aot = ee 
2 








Simplify the following by making all exponents + after 
combining like numbers: 


— 3 '— 1 Lye | be pbs 
fe VawVvb ‘ ; 32a by dy ; 
4 N/a /b . 32 a byt zt 


52x 63 x 10° 





: Vv «art 5aty\ 
—3 +3 =a OF K ‘ 
aa x WD? 
2 eS ae: ees 
Vaxby 9. (a *bic!) + (a *b*c 8). 


10. (a-2b-"c~?) . (2b) . (adic). 


190 THE ESSENTIALS OF ALGEBRA. 


Remove the parentheses and simplify : 


11, $[(ay2 ts 12. ([§ (aw + by)?t4]%)2, 
1 

y 

14. (@’—y)?, 15. (7 — yy. 


13, (wf yaa ¢Qaytpytaat 42h 
y 


It is to be noted that while (a2b2)? = 0, (a2 4. b2)2 does not 
equala+b6. An exponent is distributive to the factors, but not 
to the terms of an expression. 


(a? +b)? = a +2 02d? +B. 
(a? + b?)2 = V/a? + b?. 
Perform the operations indicated in the following. 
16. (w+ y")(@—y""’). 
17. (@ +e+1+a'+4a-)(4—a"), 
18. (a —y"*) + (#—y}). 
19. (@ —2a+a°— 2a 1)(2? a +2"), 
20. (a ?+4+ 2a%b 14 0~*)(a1+4 bd). 
21. (a°bscta?)? a (ai bt c3 x )8, 
22, (a 2b) cmt g 3) + (a3 b3ety-2) 1, 
23. (2 abe? +. 8 a~‘h-¢2) + (6a °be), 
24, (9 avi —16a°b%) + (3 aby3 + 4 abi). 
25. (2 aty t+ y a ary ee 


CHAPTER XIV. 
RADICALS, SURDS, AND IMAGINARIES. 


134. Definitions. 


Radicals. A root indicated by means of a radical sign 
is called a radical. ) 

As noted in Chapter XII the quantity under the radical 
sign is called the radicand. 

VI, V5, Va, V28 + y? are radicals, and 7, 5, a, a + ¥3 
are the radicands. 

These radicals may also be expressed in equivalent 
expressions by means of fractional exponents. ‘Thus, 


V7 = 7, 4/5 = 58, Va = a’, Ve + 3 = (23 + yy. 
The laws of algebra apply to radicals, since radical signs 


may be replaced by exponents. All the laws of exponents 
hold for radicals. 


Thus, Vab = Vavo, 
1 
for Vab = (ys = aibr =Va Vo. 
This law may be extended to any number of factors. 
Vabe = VavobVe. 
Commensurable Numbers. A commensurable number 
is one whose value may be expressed as a fraction with 


integral terms. 
Thus, 49 = 4 





2 = 98, is a commensurable number. 
191 


gi 


192 THE ESSENTIALS OF -ALGEBRA. 


Incommensurable Numbers. A number which can not be 
expressed as a fraction with integral terms is called an 
incommensurable. 

Thus, V2 = 1.4142... is inecommensurable. 

Surds. A surd is an incommensurable root of a com- 
mensurable number. 

Thus, V2 is a surd, for it is an incommensurable root of 
a commensurable number. 2 is commensurable, but V2 
is 1.4142..., an incommensurable number. 


\ 1 + -vV2 is not a surd in the sense of the above defini- 
tion, for 1 + V2 = 2.4142... is itself an incommensurable 
number. 

Examples of Surds: V3, V2, V5, V7, Va. The latter 
expression, Va, is a surd if a be not a perfect square. 

A surd is always a radical, but a radical is not always 
a surd. 

V5, Vat, V21, V16, V22y8 are radicals, but V16, V22y 


are not surds. 


135. Surds Expressed Graphically. Many surds may be 
expressed graphically. In doing this, use is made of the 
Pythagorean proposition. Jn a right triangle the square 
of the hypotenuse is ia to the sum of the squares of the 
two legs. 


If AB = BC=1, 
then AC= V1? 4 12= V2. 


RADICALS, SURDS, AND IMAGINARIES. 193 


If DEH=2and EF =1, 
then DF=V+4 2 = V5. 


If X is a cube whose volume is 5, 


then’ KL = V5. 





EXERCISES. 


Represent graphically 
dae |) 2. V3. SOuNALS. 4, V/34. 5. V6. 


136. Surd Forms. Radicals whose radicands are alge- 
braic numbers are generally considered swrds unless the 
radicand is the nth power of an algebraic number, » being 
the index of the root. 

Vat 4, Va + 6)%, ae talk ay are surds. 

V(a—x)3, Vata, V2 +22ry +7? are radicals, but not 
surds. 

These latter expressions are frequently spoken of as 
being in the surd form. 


137. Irrational Numbers. An expression involving a 
surd or surds is an irrational. 


5+V6, 34+V2—V5 are trrationals. 


194 THE ESSENTIALS OF ALGEBRA. 


138. Kinds of Surds. ‘The order of a surd is denoted by 
the index of the root. V5, V4, Vz are surds of second, 
third, and nth orders respectively. Surds of the second 
order are generally called quadratic surds, those of the 
third order ecubie surds, etc. 

Surds are of the same order when they have the same 
root index. 16, Vz, Va+6 are surds of the same 
order. 

A monomial surd consists of a single surd term. 

A binomial surd consists of two surd terms, or a surd 
and a rational term. V5 +-~V72, 5+-V10 are binomial 
surds. 

A trinomial surd consists of three terms, two of whick 
at least are surds. 3+V2—V5 and V3 4+~V2—4V5 
are trinomial surds. 

A mixed surd consists of a rational factor and a surd 
factor. 5V3, 4V2, aVx +y are mixed surds. 

A surd is in its semplest form when the root index is 
the smallest possible and the radicand the simplest possi- 
ble integral expression. 


V2T=V9x38=V9x V3=3V3 (simplest form). 
/36= V6 (simplest form). 


Noe ye =x Vab = 2 - Vab. 


Similar surds are re La when reduced to their 
simplest form, have the same surd factor. 

5V4, 8V4 are similar surds. 

In this chapter only quadratic surds will be considered. 








139. Transformation of Quadratic Surds. <A _ rational 
quantity may be put in quadratic surd form by squaring 


RADICALS, SURDS, AND IMAGINARIES. 195 


it and indicating its square root. Thus, 5 = V5? =V25. 
This is readily extended to the case of reducing a mixed 
surd to an entire surd. 


56V38 =V25 x 8=V71b; 4V2=VI16 x 2 = vB2. 


A quadratic surd is reduced to its simplest form by | 
factormg the radicand and removing to the left of the 
radical one of each pair of equal factors. 


(1) alin 2 ed = 
(2) Simplify V75 + V243 —-V108 — V27. 
V5 = VEX S =5V3. 
V243 = V2 x B= OV. 
V108 = V6? x 3 = 6V3. 
VIT=V 3x 8=3V3. 
The expression now is 


5V3 + 9V3 — 6V3 — 3V3 =5Vv3. 


EXERCISES. 
, Simplify the following surd expressions: 
fey oO 4/5 + 6-8. 
2. V124+V27 —V75. 
3, 50 —V32 4+ 2V18—5V8. — 
4. V3004+-V108 — -V243. 
5. 3-Va8y! — 2V ay? + 13-V a7. 
6. 25 a®b3 —-V/81 ab +-V144 a'd*. 
7. V8(@+ y)? —5V27 (e+ 2ay + y) +-V12(@+ y\(at+y). 
8. Vaety +a) Vo avy +2 +3VI6W +5! 





196 THE ESSENTIALS OF ALGEBRA. 


V125 + -V.245 —-V/320 + -V405 — V720. 

10. V82— 24 97+18 2 — V2 o— 1207+ 18 2. 

11. V8 m3 — 16 mn + 8 mn?—-V2 m3+ 4 mn + 2 mn? 
12. («—y)V84V32+ 6 ay + 3y?— («+ y)V108. 
13. V4a2+4a%y 4+ V4 ay? +4 9. 

14. 59 a’) +27 a? +7 aV25 64 75. 

15. V48 ay + yV75 2+ V3 2@—9 yy 

16. V75 a& — V3 a3 + 27 ab? — 18 ab + V27 ab? 

17. V5(a—b)?—V 20 a? +40 ab +20 b?+ V20 a?—40 ab+ 20 b2 
1s. 99 — -V176 + V539 + 4/275. 

19. VW52—38V117 + 5V 1578. 

20. (a+b) V(a— b+ 9) + (a—b) VE ya bp 
+(8a—4b)V8a+4b)(e+y). 


so 























140. Product of Quadratic Surds. Products of surds obey 
the following law: os sy 7) 
(1) V3iT x V388 = 38V3 x 4V9 
=12V3xv2 
—12V6. 
(2) V3 (a+b) x V2(a—b)= V6 (a2 — 8). 
(3) V150 + V48 =5 V6 + 4V3 = §V8 = §V2. 


EXERCISES. 
Multiply the following surds: 


1. 20 x V/80. 3. 5V3 x V48 x V75 x V'15. 
2. V32 x V200 x V50. 4. 3Vaxr x V16 aba x V48 aa’, 


RADICALS, SURDS, AND IMAGINARIES. 197 


. Ve@e—yyPx Ve@— yy. 
. V60 a?y? x W135 ay x V36 aty®. 
. V8(a—bY x V2(a—b) x V6 (a— bY. 
8 Vb(e—y) x V20(@+y) x V3 (e— ye +ay +). 
Simplify : 
V5 aa? x V72 a+ y) x V2(a—y) 
V25 wat x V82a(@—yra@ty) 
10. (32 + V48) + (V2 + V3). 


No o 


2, 








141. Multiplication of Polynomial Surd Expressions. Such 
expressions as 
(a+vVb+ Ve) x(a—Vb4+Ve) 
are multiplied together in the same way as_ integral 
expressions, the extended law for multiplication of radi- 
cals being observed. 


(1) a vb +) Ve 
Fee MN Ds tt ENG 
ataVvb+ ave 
Say b ab — V/ be 
eat + Vbe+te 
a? Pare —b +e 


(2) +v3) x (V2— V5)=v24+V6—V5—VI15. 


EXERCISES. 
Multiply the following: 
1. (1+-V3) x (1—-v3). 5. (3V2 —4)(8V2 4-4). 
2. (2—¥V5) x (2+ V5). G0 V,0 — 2a hi 
3. (V24+V3)(V2— V3). 7. .(V2-+- V5). 
4. (2V38—1)2V3+1). 8. (V2+V3—1)% 


198 THE ESSENTIALS OF ALGEBRA. 


9. (2V3+1)(V3 + 2). 

10. (V5+V3+4)(V5—V3—4), 

1. (2V3—3V5+1)(5V3 +2V5—1). 

12. (4-7 — IV 6) Ta 6 

13. (V3+1)(V9—V3+1). 

14. (Vat Vb— Ve)(Va— Vb + Vo). 

15. (2Va—3Vy —5-V2z)(2Va+ 8Vy —5-V2). 
VbB-V7T_ , 2V54+2V7 

2V3—-4V5 8V346V5 


142. Conjugate Binomial Surds. 
Two quadratic binomial surds differing only in the sign 
of a surd term are called conjugate surds. 
Va+tvb and Va—-vob, or 4+ Vy and 2—Vy, 
are conjugate binomial surds. 
(Va+vb) x (Va—Vb)=Vai—-VP=a-5. 
(x+Vy) xX (@—-Vy)=2—-VyP= 2 —y. 


The product of two conjugate binomial surds is rational. 


143. Rationalizing Factors. When the product of two 
surd expressions is rational, one expression is said to be the 
rationalizing factor of the other. 


For example: Va— -Vé is the rationalizing factor of 
Va+vb, because (Va+Vb)(Va—Vb)=a—8, a rational 
result. 

a+vVy is the rationalizing factor of 2— Vy, because 
(@—Vy)@+Vy)=v—y. 

V5 is the rationalizing factor of V5, because V5 x V5=5. 


RADICALS, SURDS, AND IMAGINARIES. 199 


Suppose the value of is required. It will 


re 
V5—VB 
evidently be very much simpler if we first rationalize the 
denominator. 











V5. - V5(V5+-V3) _5+V15_5+4+V15- 
V5—V3° (Vb—V3)(Vo+V38) 5-38 
pe V2 VT V2) V2) VIS 
meee 69 i -Vvav2. 2 
EXERCISES. 


Rationalize the denominators of the following fractions: 











1. ah 5. fal ee —s 9. 2 a V3. 
v2 42/3 ay ie 
2. L+v2. 6. pee Vie —* 10. 3 ds v5, 
V3 Bo 1G hee sss 
eg, SVE nn, 442 V3. 
2V2 4—3~V7 V5 —V3 
4 fp ae Taye = aoe 
13. 5+8V5 | a7, Lt 8ve—1. 
2V5—V3 1—3Vae—-1 
14 84+2V2+V3_ 18. OV GAEN Fs 
2V543V7 (@+y)+V2ay 
15: Set ilp aby a—Vayt+y. 
Va+ Vb Vae—wVy 
Tee PVE 20, Ve EN 2 ae 


2Va+4vb ' Va—2b—-V2a—b 


200 THE ESSENTIALS OF ALGEBRA. 


144. Rationalizing a Trinomial Surd. A trinomial surd 
expression may be rationalized by two operations. 

(1) Rationalize Va+Vb+~Ve. First multiply by 
Va+tvb—vVe. This gives a+6b—c+2Vab. Now 
multiply bya+6—e—2-Vab. This gives (a+b—c)?—4 ab, 
a rational expression. Hence the rationalizing factor of 
Vat+vb+Ve is (Va+vVb —Ve) (a +6 —e — 2 Vab). 

(2) To rationalize V2+V38+~V5, we multiply by 
(V2+V38 —V5)(2+38—5—2-V6). This will give a 
product 24. 

EXERCISES. 


Find the rationalizing factors of the following: 


Dee tny Cerys 5 VW1i0=V2se 
Den) SaNy & ANT 6. Va+vVore 

SINS +V 24/0. 7, 1—2 Voeee ee 
Bey Gey ols 8. 2Va—V2b4+3Ve. 


145. Rational Numbers and Surds. 


THeorem I. Jf Vax and Vy be surds, then Vx can not 
equala+~Vy, where a is rational. 


For, assuming Vz=a+-Vy, and squaring, we have 
e=a+y+2a Vy; 


x—-@—y _ 
or oS Gir 


But the left member of this supposed equality is rational, 
and therefore can not equal the surd Vy. 


Hence, Vaetat+vy. 


The sign + is read, is not equal. 


RADICALS, SURDS, AND IMAGINARIES. 201 


THEOREM IT. ais a +Va= b +Vy, where a and 6b are 
rational, and if Vx and Vy are surds, then a=b and x= y. 


The proof of this theorem is similar to that of Theorem I. 


We have te Vie 

Squaring, (a—6)?4+2(a—b)Vr+ae=y. 

Transposing, — x+y —(a—b)?=2(a—6b) Vz. 

Here, a rational number, —x + y —(a— 6)?%, is equal to 
2(a—b)timesasurd. But this can not be true, except in 
the case a=0b; but when a=6 the original equality says 
that x= y. 


146. Square Root of a Binomial Surd. ‘The square root of 
certain binomial surds may be extracted. 


(1) Find the square root of 5 + 2 V6. 


Let V54+2V6=Vr+ Vy. 
- Then §+2V6=ex+y4+2Vz2y, by squaring. 


ey =O, ry ==, by Theorem II. 


The question now is to find two numbers whose sum is 
5, and whose product is 6. ‘These are seen to be 2 and 3. 


Hence, V5429V6= V3 4 V3. 
(2) Find the square root of 8 — 2 -V15. 
As in (1), let © V8 — 2/15 = Vz— Vy. 


Then 8—2Vib=2+y—2Vz2y. 
r+y=8, vy=15, 
then sie py 


Hence, V8—2V15 = V5— V3. 


202 THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 


Extract the square root of the following: 


Tent Oise won 7.31.2-e a 

Qa iee wey oo 8. 1824742, 
Beeliletaoeny Og 9. 49 +125. 

4. T—V40=7—2-V10. 10, ¥—-2V yee 

5, 1642/55. 11; 87 — 12Ra 

6. at+b—2Vab. 12. at+b+c+2Vac-+ be. 


147. Imaginaries. An imaginary number has already 
been defined as the even root of a negative number. 

We shall have occasion to use only the square root of 
negative numbers. 


V/—5, V—10, VEo16 =4V— 1 eee 


are imaginaries. 

It is to be noted that every square root of a negative 
number may be expressed as a real number multiplied by 
the square root of —1. Thus, 


V—16=V16 V—1=4V_-1; V—5=VvV5 V-=—1; 
V—@=VaiV—1=avV—1; V—a=Va V1. 

The V—1 is usually denoted by 7, and is called the 

imaginary unit. 

148. Some Properties of / =V—1. 

(1) i =V—1. 

(2) 2=(V—1)?=~-1. 

(3) &®=(V—1)8=71() = -7. 


RADICALS, SURDS, AND IMAGINARIES. 203 


C4) = (2)? =1. 
(5) Bait) at. 

(6) 6-1) =2= —1. 
(T) Tat) =-2. 
(8) B=(i)2=1. 

(9) =i) =1; ete. 


This table shows that the powers of ¢ repeat the values 
V—1, —1, —V-—1, 1, in cycles of four. 


149. Operations with Imaginaries. All the operations 
possible with surds are also possible with imaginaries. 
The properties of 2 must be observed. 

(1) V—36 +V—81+v-— 100 

= V36(— 1) + -V81(— 1) + V100(— 1) 
=6V—14+9V—1+10V—-1 
=67+97+102 
e200. 
(2) V—48 + V— 75 +V— 248 
= V16(— 3) +-V25(— 8) + V81(— 8) , 
= 4iv8 4 5iV3 + 9iv8 
SS B/S. 
(8) V—5xV—T=Vbixvit 
=V5VT ii 
= V35? 
= — V35. 


204 THE ESSENTIALS OF ALGEBRA. 
(4) (V—84V—2)(V—8-—V—2) 
=(iV3 +iv2)(iv8 —tv2) 
= 12(V3 + V2)(V8 — v2) 


. = 12(3 — 2) 
= —1(3—2) 
Lies Boia, 
(5) 8+iv2)?=94 6iV2472-2 
—-9+6V2i-2 
—7T+6V24. 
5 4V37 
CDE TAY Ey 


To rationalize and make real the denominator, multiply 
both terms of the fraction by 5 + -V37. 


Oe — ee 


(5 —V3i) (6 +V3t) 25 — 37? 25 —(— 3) 
_ 224+10V32 
er anes 
_ 114+ 5V3¢ 
14 
EXERCISES. 
2, V—18+V¥—128 — VS bt 
2. (31+V21) x5V2i=? ¢ StV=5_» 
3. (4—V3iy=? 38—V—5 
4, (V—34 V2) (V—243)=? 7 22a 
PEED Su le: 34+ 2V3% 
V3—-V—2 8. V—75 +V—25 =? 


(Make denominator real and 
rational.) 


RADICALS, SURDS, AND IMAGINARIES. 205 


9. (av —a2+yV— y) (av —2—yV—y) =? 





Tig =» sir Rice ae 
a— bi V/—5x 





hg BSL L234 
2 — i WS eel ee 
“(Oia | Cormar 


150. Graphical Representation of Imaginaries. Complex 
Numbers. We are accustomed to represent real numbers 
upon a straight line, the positive numbers in one direction 
and the negative numbers in the opposite direction. 


Let OA =1, 
OA' =—1, 
1x2=—1. 





Hence, multiplying 1 by 2? turns it so as to reverse it in 
direction. OA is turned about O to the position O0A' when 
it is multiplied by 7?; that is, OA is turned through a half 
circle. From this we conclude that, when OA is multi- 
plied by 2, it will be turned half as far; that is, to the 
position OB. If OA is multiplied by 23, it will be turned 
to the position OB’. It is customary to regard BB’ as 
the line of imaginaries. 

A number like 3+ 27 is called a complex number. The 
type form of such a number is 


a + 67. 


2.06 THE ESSENTIALS OF ALGEBRA. 


The following diagram shows how complex numbers are 
graphically represented. 


(P,) -2+22 (P) 3424 
@ e 





EXERCISES. 


Locate on a diagram the following complex numbers: 


1. 447. 4 —347. 
2. —3— 31. 5 — 34. 
3. 2—5%. 44¢ 


CHAPTER XY. 
QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 


151. Definition. An equation in which the highest power 
of the variable is two is called a quadratie equation. 


Thus, 327+427+5=0, 2-—5274+1=0, az’?+6=), 
are quadratics. 


152. Type. Every quadratic in a single variable may 
be reduced to the type ax?+ 6x+e=0, where a may be 
any quantity except 0, 6 may be any quantity, and e may 
be any quantity. | 

3a2—4a2+5=0 is a special form of the type, in which 
a=3, 6=—4, ande=5. 


EXERCISES. 


Reduce the following equations to the form of the type 
ax’ + 6x +c=0, and indicate the particular values of a, 6, and 
c in each case: 











IS a = o — 5. ee ae 
( Ppa 
2. 6a2=2—27. 
: , 5 t—4_7a°-3 
(Clear of fractions.) Mice Tinka ° 
2e—T 2—27 1 3 
3 ———_ = —_—_- 6. x©+-=5—--.- 
3 5 % 


207 


208 THE ESSENTIALS OF ALGEBRA. 





xv? — 4. 3  8e+1 
7. —— =2 =, . ——=22. 
oq ala 9 z 22 
8. iO a Hag) 10. da+1 2042. 
e+ 2 e+1 e24+2 


153. The Pure Quadratic. If 6—0, the general quad- 
ratic ax?+6x+e=0 becomes ax?+e¢=0, an equation 
sometimes called a pure quadratic. 

The solutions of ax?+e¢=0 are easily found. 


Transposing e, ax*=—e. 
Dividing by a, a= —*. 
a 


Extracting the square root, z= a 
a 


If a and ¢ have like signs, the roots +{—-— are both 
a 


imaginary ; if a and e have unlike signs, the roots are both 
real. =e 
Thus, ose = 0 has the roots r= oe ee vee are 


imaginary; 827—5=0 has the roots 7= ssf , which are 
real. 


The roots are real and rational if a and ec have unlike 


e Cc e e . 
signs, and — is a perfect square; the roots are irrational 
a 


C .e 
when — is not a perfect square. 
a 


Thus, 


322—5=0 hastherootsra= + , which are irrational ; 


Se 


| 


| 2) 


Pe +3, which are rational. 


3a2—27T=0 has the rootsz = wp 


Se) 


~ 


QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 209 


EXERCISES. 


Reduce the following equations to the form of the type: 



































| ax’+c=0. 

Mee fe 0. Gee eee 
7 3 32 4x4+5 
(Clear of fractions.) 2 S2t+4_ 5a—3 

G—-3 _ ea pat Se oi 4 

5 U+9 1 P 
ee er g, =" =" _. 

3. = = 
7 142 ; 2 eae 

” ia © 9. e,4_5 
ed @+3 hia el 
ne ee 6 v 10. Sal ea be % ° 

72 11 x 3u+7 


EXERCISES. 


Solve the following equations : 


ie 7e?—112=0. 
First SoLuTIon. 

Transposing, ee 12, 

Dividing by 7, re 1G: 


Extracting the square root, x= +4, rational roots. 


SECOND SOLUTION. 

The general pure quadratic 
ax?+¢c=0 ‘(uh 
has the solutions r= Ly = “ 


Comparing 7 2? — 112 — 0 with this general equation, 
ant. eo = = 112, 


Ni 1155 fii2 


210 THE ESSENTIALS OF ALGEBRA. 


























c—3 12 6.00 = eee 
2. se . YP cig =-—2. 
5 ees te a 
x — Doe goo Oe 49 
BA ————— eo . 38e __3a+30 
7 14 2% 8. 9b ee 5 
4. (w©+3)(@—3) =2a?— 45, 
5. 3¢—27=721. 9. of 4+ 6 ee 
4 3 3 
6 OX Se! 
es eee 10. —aa’?+25b=0. 


154. Solution of the General Quadratic, ax?+ 6x +¢=0. 


Dividing through by a, we have 


b 
v2+—-—x rin ett 
a 
my « ~ Cc 
lransposing, e+ -_7 = 
a a 


2 
Adding + (the square of half the coefficient of 2) to ° 
t 


both sides, we have 
b b? igh en 
7 aes ee 
—4 ae 
4 a? 
This is called completing the square, because it always 
makes the first member of the equation a square. 
Iixtracting the square root of both sides, 


b Nees: 4 ac 


C--— = 
2a 2a 
arising: Vh2— 4 ae 
ransposing, Gaon A” ae a 
2a a 
—b+v82—4 ae ac 


2a 


QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 211 


_ 


This result shows that the quadratic ax’ + 6x +¢=0 has 
two roots; namely, 


ya b+ ve— Fae 
2a 

a Pen vO pe aC: 
2a 


This is the solution of the general quadratic and may 
be taken as the type solution for all quadratics. 


Solve 327°+827+5=0. 
First SoLurion. 
Dividing by 3, a+ $r+3=0. 


Transposing, a+ bas — 


color 


Completing the square, 


at +4. 8 ax + 16 = 16 — 





Extracting the square root, ° 


=— land —8. 
Verifying the Solution: Put x=—1 in the equation 
3a°+82+5=0 and 


pea yt8 (1) b= 8 — 845 — 0. 


Put s=— 3 and 


3(— H+ 8(-H4+5=%— 4045 = 4h 40 4 = 0, 


3 3 
Replacing the variable of an equation by a root reduces 


the equation to an identity. This process is called verify- 
ing the solution, 


212 THE ESSENTIALS OF ALGEBRA. 


SECOND SOLUTION. 
Comparing 322+ 82+5= 0 with az7+ br+e¢=0, 
a=8, b=8, and c= 5d. 
Let us substitute these values in the type solution, viz., 


—btive?—4ae 
= raat PEREIRA Se ara | 
Qa 


= 8 4 vi64— 60 
a | 
_—8+v4 
Seren: 
—8+2 
6 
It will be noticed that the substitution in the type solu- 
tion gives results identical with those obtained by carrying 
out all the steps of dividing, transposing, completing the 
square, extracting the root, and transposing. 


and we have L£= 





=-—iand —§. 


EXERCISES. 


Solve the following equations by both processes in the order 
of the above illustration and verify your results ; 

1, 22?7—52=3. 10. 11 2? =: 39 2 4+ 20. 
2. 42°4+7x¢= 16. ll. 622+72—20=0. 
3. da’ =19 e+ 14. 12. 6a°—47 e+4+77=0. 
2 5 o* —12 = 4a. 13. 3¢7+427+10=0. 
G 


. 63"+¢—156=0, : 


ene 5+ 18. 2 

15. 3%——=1. 
ve Si a ie 2 
&. 150?—8= 37a. 1s. 1+ f2=—3 28 
9 777+7=502. 17, ve’ +1=-—2.9¢24 





QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 


‘ 



































1 3 8 12 
18. “a nee =—-e 20. ie | es ne 
e—-2 #e2+2 5 x re 
21 —38 80 
[| . a 2 —— =0 
5x +20 20—5a ee 
x x—3 x 2 os at ae 
22. -— ig el eee SS ee 
e—3 x bake x 3 
23. 2+2ba+c=0. 
24, 2 —(m+n)e+mn=0. 
25. (a—b)x*—br=a. 
26. (a*— b*)x’ — (a’+ b*)~r+ab=0. 
7 
b—a b+a V—x’ 
28. Ve—5ae= eo 
Ve—5e2 
29. am poo 
Vie 827+ 5 
30. Vae+ca= gene —— 
aa” +- cx 


155. The Double Root. 


Solve 


Comparing with the type az? + br+ ¢=0, we have 


Hence, 


9227-2074 50=0. 


a= 2, 6=— 20, e= 50. 
20 + -V400 — 400 
a 
4 
_ 20+V0 
ea. 
=5+0. 


5) 


od 


13 


914 THE ESSENTIALS OF ALGEBRA. 


In this case the roots are 5+ 0 and 5—0; that is, 5 in 
each case. Hence, the equation has two equal roots. In 
the above example 5 is said to be a double root. ‘The 
quantity under the radical, 62—4 ae, is 400 — 400 or 0. 
Hence, the condition for a double root, or two coincident 


roots, 18 
; b?— 4ac=0, 
where a, 6, e¢ are the coefficients in the general quadratic 


az? + br+c=0. 


EXERCISES. 


Solve the following equations, noting those which have double 
(coincident) roots : 


1. wo —4e44=0. 6. #—le+7=0 
2. 4e¢?7+424+1=0. 7. 5° —4 9 aw 

6 HS SUF et elt). 8. 2527+ 302+9=0 
4. 4e?—127+4+9=0. 9. fv’+ie«+1=0 
5. 8e?+42+1=0. 10. —3e¢’+42+:=0. 


156. Irrational Roots. 
Solve 3822? -—-927+4+2=0. 
Comparing with the type 

a+ br +e=0, 


we have a= 3,6=—9, e= 2. 
Hence, r= 94 V81— 24 
6 
_9-v51 


6 


QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 215 


Here it will be noticed that the roots are irrational. 
The quantity under the radical, 6? — 4 ae, is 81 — 24 or 57, 
and is not a perfect square. ‘he roots are conjugate surds. 


EXERCISES. 


Solve the following equations : 














1. 42? +2—1=0. 5. Bgl Vee Wad 
3 5 
2. 3x —Te#+3=0. 6. (v—1)(a—2) =1. 
Se te — 7 — 0. 7. 807 —21=202. 
oe: ike ees 
4. 52° —32—3=—0. BOGS pate ae 
> Siig see ea licas see 
2 D 2 \ 
10. 9 m'n*a? — n? = 6 m?n®a. 
157. Complex Roots. 
Solve 627 —Ta+3=0. 
Comparing with the type 
ax? + br +e¢=0, 
we have a=d,6=—T,e¢= 38. 
T+V19 — 60 
ae Pt Vette 00 
10 
_T+v—11 
10 
_T+vil-2 
LO aie 


In this case the roots are imaginary. The quantity 
under the radical, 6? — 4aec, is 49 — 60 or — 11, which is 
negative and therefore its square root is imaginary. ‘The 
roots’in this case are conjugate complex numbers. 


216 THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 
Solve the following equations : 
1. #—52+8 —0. § —327+11382=20. 
2. 22°+924+11=—0. 6. 2? + .54¢74+ .3=0. 
3. 3a?—10724+9=0. 7 — 32° 4+.82 —10= 0) 
4. Te—l11v7+8=0. 8 


158. The Discriminant. The solution of the general 
quadratic ax2 + bx +¢=0is 


Pater seh A Nee et 
2a 


From the examples just given we see that the character 
of the roots is determined by the quantity under the radi- 
cal. This quantity, 62 — 4 ae, is called the discriminant of 
the quadratic. 


159. Some Conclusions. 


(1) When 62 —4ac =a square, the roots are real, rational, 
and different. 

(2) When 62—4.ac=0, the roots are real, rational, and 
equal. 

(3) When 62—4ac =a positive number not a square, the 
roots are real and conjugate surds. 

(4) When 62—4ac=a negative number, the roots are 
conjugate complex numbers. 


(1) 4a -—Tr+38=0. 
The discriminant 62 — 4ae= 49 — 48 = 1. 


Hence, the roots of this equation are real, rational, and 
different. 


QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 217 


(2) 2a7?-—4742=0. 

The discriminant b?— 4 ace=16—-16=0. 

Hence, the roots of this equation are real, rational, and 
equal. 

(8) 5a74+82—2=0. 

The discriminant 6? — 4 ac = 64 + 40 = 104. 

Since 104 is not a square, the roots are conjugate surds. 

(4) Tz?-—5274+1=0. 

The discriminant 0? — 4 ac = 25 — 28 = — 3. 

Since the square root of — 3 is imaginary, the roots are 
conjugate complex numbers. 


EXERCISES. 


By means of the discriminant tell what kind of roots belong 
to each of the following equations: 


a2 


a » 2 


e—d5ae—9—0. 6. 
(7 +32—1=—0. 7. 
5e7+9e2+11=0. 8. 
62?—-Ta+$=0. 9. 
9e?—1827+4=0. 10. 

EXERCISES. 


ax’ + 2aux--(a—4)=0. 
1le’?—32e4+7,=90. 
e—3ae+$=0, 
4¢+132+11=0. 
ax’? +5e2—1=0. 


Solve the following quadratics by comparison with the solu- 
tions of the type: 


i 


ee i net 


$e +5a2—3=0. a 
327—62+1=9. 8. 
—227+82+6=—0. 9. 
6a? +a—2=—0. 10. 
24¢7+142—5=0. 11. 


xv” — 55a —.065 = 0. 12. 


5 a? — 500 = 0. 
627+ 82=0. 
92? — 252+ 50=0. 


307+ 212—5=0. 
7e?’—424+3=0. 


918 THE ESSENTIALS OF ALGEBRA. 


13. 1a°’-—6a%—13=0. 15. lla’—2a4+7=0. 
= (), 16. jv —5248=0. 


160. Graphical Solution of the Quadratic. The general 
quadratic ax ti bx t¢e= 0 
is equivalent to the two equations 


Y= ur, 
ay+ber+e=0. 

For if in the second of these, we put the value of y from 
the first, we get the quadratic aa*+ br +¢=0. We know 
that y + bx + ¢=0 has a straight line for its graph. (See 
page 1538.) 

Let us see what the graph of y = 27 is. 

y can not be negative, because a square can not be 
negative. 

Solving for 2, we have 








Y a=tvy. 
ageoy) Tk y=0, «=0. 
y=1, x=+1and —1. 


(4,16) 2, w=+1.414 and —1.414. 


3, v©=+1.7382 and —1.782. 
y=4, x=+2 and —2. 
etc. etc. 


(314,121%) 


(3,9) 
Representing these points with refer- 


(24,6'4) ence to the codrdinate axes and draw- 
ing a smooth curve through them, 
(1,24) we have the curve of the adjacent 
- figure, which is the graph of y= 2%. 
This curve is a y-parabola and is 
the same for every quadratic. 





QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 219 


The line az+ by +e¢=0 can only be specifically repre- 
sented, when we give particular values to a, 6, and e. 

If the quadratic becomes particular, then the lne is 
specific, and we can readily draw its graph. 

Determine by means of graphs the roots of 


v—x—2=0. 
This equation is equivalent to 


y = x", 
eo, 
In this a=1,6=—1, ande=—2. 
In the equation y—x—-2=), 
when y= 0, x=— 2, and when z= 0, y=2. 


Hence, the x-intercept is — 2 and the y-intercept is 2. 

Now, drawing the graph upon the same diagram that 
contains the y-parabola, we get the result Y 
shown in the adjacent figure. 

It is seen that the line cuts the y-pa- 
rabola at P and Q: The z of P is OY, 
which is — 1, and the 2 of Y is ON, which 
is’ 2. 

The roots of the quadratic 72—2—2 
= 0 are —1 and 2. 

Hence, we see that the intersections of 
the line az + by+e= 0and the y-parabola 
y =x" have for their abscissas the roots of the quadratic 
ax? + 6x+ec=0. 





161. Graphical Solution of the Pure Quadratic. If the 
quadratic is az?-+¢=0, the two equations to which it is 
equivalent are y = 2, 


ay+e=0. 


220 THE ESSENTIALS OF ALGEBRA. 


Here again we have the same y-parabola. The line ay + 
c=0 is parallel to the z-axis, and so the abscissas of the 
two points of intersection will be equal 

in value but opposite in sign. 
Determine by means of graphs the 


P Q roots of Pee Pig ee) 
The equivalent equations are 
(ia 
The: graphs show the two roots to be 
+2and —2. 
162. Graphical Solution in Case of Equal Roots. 


Construct graphs showing the roots of 


v—47+4=0. 
The equivalent equations are P 
oa 
y—4x24+4=0. 


In this case the line just touches the 
parabola at P, whose abscissa is 2. 

The quadratic z7—4 7+4=0 has equal 
roots, each of them being 2. 


163. Graphical Representation in Case of Imaginary Roots. 

Construct graphs showing that 27—2a2+5=0 has imagi- 
nary roots. 

The equivalent equations are 


eink 


y—224+5=0. 


QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 221 


This equation has imaginary roots. The line 
y—2x4%+5=0 
does not touch the parabola y = 2%. 


164. Certain Conclusions. 


(1) In the case of real different roots, 
the line ay+x+e=0 cuts the parabola 
y = x" in two places. 
(2) In the case of real equal roots, the 
line ay+bx+ec=0 touches the parabola 
y=x", but does not cut it. 
(3) In the case of imaginary roots, the line ay+bz+c=0 
is entirely outside the parabola y = 2. 


EXERCISES. 


Using the parabola in the book, or a similar one of your 
own construction, solve by means of graphs the following 
equations : 


1. 2¢+4¢—-1=0. 

Norr. The line in this case is2y+2-—1=0. 
fA yf aards 
ee se 


gz and y intercepts are 1 and }. 

Merely lay a ruler across the parabola so as to 
make these intercepts, and note the abscissas of 
the points of intersection with the parabola. 


2. @—S32+2=0. 6. 52?—6a—8=0. 
3. 4e¢7+4274+1=0. 7, #—Azgt4d=0. 

aoa O80 4-0 = 0, 8. 2307-524 3=.0, 
5. 102°—32—4=0. 9, 27—424+3=0. 


ph THE ESSENTIALS OF ALGEBRA. 


165. Relations among the Roots and Coefficients of a 
Quadratic. The roots of aaz2+bx+c=0 are 


_ —b+VP—4a0 fae 
AE, 

ray ly clase mad) tac 
2a 


Here we use « (alpha) and ® (beta) to represent the 
two roots.. 
Adding these two roots we have 
—~btVRPo dae 2b = Vee 


1 eo ai a 


Multiplying together these two roots, we have 


yaw alge ou kabey Ady, 2 2 
of = (= +5" =a\ 6—vb? at) 


2a 2a 
VC eine: Cokes 
4 a? 402 a 
If the coefficient of 27 in the general quadratic 
av?+ br+e= 0 


be made unity by dividing by a, the equation takes the 
form 


b é 
g+—71+—-=0,. 
nie 
The sum of the roots e+ B=— a and the product of 
| a 
the roots «8 = <. 
Hence, if a quadratic be written so that the coefficient of 
x is unity, the coefficient of x is the negative of the sum 


of the roots, and the constant term is the product of the roots. 
This has already been seen in factoring, page 86. 


\ 


QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 223 
The roots of the equation 25a’—15a%+2=0 are 2 and t. 
When this equation is divided by 25, it becomes 

o—se+7,=0. 


The sum of the roots is 2+ 4= 3, and their product is 2. 
This sum and product are, respectively, the negative coefti- 
cient of # and the constant term. 


166. Formation of Equations with Given Roots. If the 
roots of an equation are 2 and 1, what is the equation ? 
Here 


and 


Colbo 
~ 


| 


| 
coi bo ke 


g 

on 

jes teae = 
I| 


or 38a2—5xr4+2=0. 
In general, 227—(a+8)x+e8=0 is the quadratic 
equation whose roots are « and £. 
EXERCISES. 


Make equations which have the following roots: 


1. 3, —4; —5 4; 3,4. 6. 3—iV2, 3+iv2. 
2. 1 +~/5, 1 Lin/5. wh 3 — 5,V5, 8+ 5,5. 


e354, 8— bi. 


8. a+bi, a— bi. 
ae es 9. 314 5Vmi, 31—5V mi. 


5. 5, — Ad. lo. V5+3V3, V5 —8v3. 





224 THE ESSENTIALS OF ALGEBRA. 


167. Generalized Quadratic. aX?4+6X+c=0. 
The above is a quadratic in X. Its roots are 
—b+VP—4ae 
2a 

X may be any algebraic expression. Whenever an 
equation may be arranged like the above type, it is said 
to be of the quadratic form. 

C1) 324+797+4=0. 

Put 2? = X, and the equation becomes 

38X274+7X+4=0, 


xX = — 








whence X= sieve. 
=—l and — 4. 

But X = 7, 
hence, a= —1and — 4, 
and g=+2and +2V32. 

The roots of 304+ T72+4=0 
are Tivewi tore V3 4. 

(2) (a? — 5)? — T(a@#— 5) 4+ 12=0. 


Put 2 — 5 = X, and the equation becomes 
X?—-7X+12=0. 
CXa Ay XE SNe Ol 


X = 4 and 3. 
Hence, z?—5=4 and 3. 
z2=9 and 8. 


e=+3and +2v2. 


QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 225 


(3) 202 +5 Vo? —5274+3=1074+ 6. 
‘Transposing, 222—102+ 5 Va?—52+3-—6=0. 
By adding 6 and subtracting 6, we have 
9a? —107+64+5V22?—52+3-—6-—6=0, 
or, 0 A(a@®' — 524+ 8)4+5V22?—524+8-—12=0. 
Put Va? —52+3=_X, and the equation becomes 
2X24+5X—12=0, or (2X —3)(X + 4)=0. 
X =— 4 and 3. 
Since X = V22— 52+ 3, we have 
Ve—5x2+3=—4 and 3. 
a—5x2+3=16 and 2. 
v—d5xe—13=0 and 2—52+3=0. 


_5+V25 + 52 54VIT 


b) 





2 2 
Fav 8) ba! 
and ee ee ea a 


The roots are 5tViT and. 5+ V22. 
2 2 
EXERCISES. 

Solve the following equations: 

1 (4+32)4+38V#+3e—4=0. 
(a? — 2)? — 5 (a? 22) +6=0. 
e828 oD & 0. 
Vb 02+ 404+3(5a2+40) = 24, 
(80+5)+4V32+54+7=0. 


ole died ae 


226 THE ESSENTIALS OF ALGEBRA. 


6. (2+5)+3(2+7)-4=0 
x 2 
97. 8(22+4)—4V2e+441=0, 
8. 28+727—18=—0. 
9, w+4e?4+4+4+5(a°+2)—6=0. 
lo. 28— 5a? = 36. 
11. v'—92? = 400. 
p i 2a—84+V2e—5=—1. 
13, x—5—V22—11=8. 
14. po 0 ea On/m one eee 





15. PtatvVvetet5 =25. 
16. +504+4=5V2e'?4+5a+4+ 28. 
17, @4+5e—Ve+5a+14= 42. 
ia 3 VatT_ Vets. 

Ve4-3 V2 
19. V2i—3 Va =40. 
20. 22°9—324+6V22—324+2=14. 








EXERCISES. 
1. One half a number plus the square of the number is 150. 
Find the number. 


2. The sum of two numbers is 15 and their product is 56. 
Find the numbers. 


3. Find two numbers whose sum is 80 and whose product 
is 216. 


4. Separate 41 into two parts such that the product of the 
part is 530. 


QUADRATIC EQUATIONS IN A SINGLE VARIABLE. 227 


5. Two numbers differ by 3, and the sum of their squares 
is 225. Find the numbers. 


6. The sum of the squares of three consecutive numbers is 
434. Find the numbers. 


7. A rectangular lot is 32 feet longer than it is wide. It 
contains 13200 square feet. What are the dimensions ? 


8. A man buys a certain number of chickens for $6. If 
he had paid 10 cents apiece more for each, he would have 
gotten 5 fewer for his money. How many chickens did he 
buy ? 

9. Find a number such that if its square be diminished by 
1, } of the remainder is 18 more than 10 times the number. 


10. There are two numbers whose difference is 8. If 540 
is divided by each of these numbers, the difference of the 
quotients is 18. Find the numbers. 

11. One side of a rectangle is 7 feet longer than the other 
and its diagonal is 13 feet. Find the area. 


12. The difference of the reciprocals of two consecutive 


numbers is ;34,5- Find the numbers. 


13. The difference between the reciprocals, two consecutive 


odd numbers, is x25. Find the numbers. 


14. By increasing his speed 1 mile an hour a man finds that 
he takes 5 hours less than usual to walk 60 miles. What is 
his ordinary rate ? 

15. The larger of two pipes will fill a cistern in 6 minutes 
less time than the smaller. When both pipes are open, the 
cistern is filled in 131 minutes. Find the time required by 
each pipe to fill the cistern. 


16. A and B have a distance of 150 miles to travel. B 
starts 10 hours before A and arrives 10 hours after A. A 
travels 2 miles an hour faster than B. What is the rate of 
each per hour ? 


ete maine THE ESSENTIALS OF ALGEBRA. 


17. A rectangular field is 4 times as long as it is wide. If 
the width is increased 20 rods, its area is doubled. Find the 
area of the field. 


18. What number increased by 4 and squared is equal 4 of 
itself increased by 10 and squared? 


19. A man sold a horse for $144, thereby gaining as many 
per cent as the horse cost him dollars. What was the cost of 
the horse ? 


20. A boat’s crew can row 9 miles down a river and back 
in 4 hours. The rate of rowing in still water is double the 
rate of the current. Find the rate of rowing and the rate of 
the current. 


21. The hypotenuse of a right-angled triangle is 5 feet 
longer than the base and 10 feet longer than the perpendicular. 
Find the sides of the triangle. 


22. The sum of two numbers is a and their product is B. 
What are the numbers ? 


23. The perimeter of a rectangular field is 168 rds. and 
its areais 9 A. Find the length of the sides. 


24. Two men start at the same time from the vertex of a 
right angle and walk along its sides at the rate of 3 and 4 miles 
per hour, respectively. In how many hours are they 50 miles 
apart ? 


CHAP THR @X0V I: 
SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 


168. When simultaneous equations involve quadratics, 
they must be solved by methods which depend upon the 
form of the equations. Various methods are shown 
under the following cases: 


GASH 1. 


169. A Linear and a Quadratic. A simultaneous set of 
this kind may always be solved. The equations are of the 
forms, 

(1) az?+ bry +cy=d, 

(2) le+ my = k. 


(3) se nai, from (2). 


(4) Cae Amu Tey Ea") y +cy2=d, 


by substituting in (1). 

(5) Cam? — blin + cl?) y? + (blk — 2. akm)y + ak? — di? = 0, 
by rearranging (4). 

Equation (5) is a quadratic in y, and therefore has tivo 
roots. The substitution of these roots in (8) will give 
two values of x. Hence, the set of equations has two 


roots, and only two. 
229 ° 


230 THE ESSENTIALS OF ALGEBRA. 


In the above we have used general equations and we 
have found that the solution depends upon a quadratic in 
one variable. Such a quadratic can always be solved. 
. Hence the simultaneous set can always be solved. 


I. Solve the following equations: 


(1) a? + y? = 25, aa} 
(2) Ty—a2= 26. 
(3) z= Ty — 26, from (2). 


(4) 49y?— 350 y+ 625+ y?= 25, by substituting in (1). 
(5) 50 y2— 3850 y + 600 = 0. 


(6) y2—Ty+12=0, by dividing (5) by 50. 
EO) (y—3)(y—4) =), by factoring (6). 
(8) y = 9 and 4. 
Co) 2a=7Tx38—25 and 7 x 4— 25, 
by substituting in (8). 
(10) x=—4and 3 


The roots are (— 4, 3) and (3, 4). 

Care should be given to the proper association of the 
values of x and y. It should be remembered that a root 
is a properly associated value of 2 and of y. 


The graphs of 27 + y? = 25 and T y—# =25. 
From 2* + y* = 25 we have 
y= +V 25 — 2%. 
When z= 0, y=+5 and —5; 
(0, 5), (0, —5) are roots. 
When g=+1l and —l,y= 2/6 and —2V6; 
(1, 2V6), (1, —2V6), (—1, 2 V6), (—1, —2-V6) are roots. 


SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 231 


When «= 2 and —2, y=V21 and —v21; 
(2, V21), (2, —-V21), (—2, V21), (—2, — V21) are roots. 
When «=3 and —3, y=4 and —4; 
(3, 4), (8, —4), (—3, 4), (—3, —4) are roots. 
When z= 4 and —4, y=3 and —3; 
(4, 3), (4, —3), (—4, 3), (—4, —3) are roots. 
When x=5 and — 5, y= 0 and 0; 
(5, 0) and (—5, 0) are roots. 
Locating these roots and drawing a curve through them, 
we find the graph to be a circle. 
The graph of Ty—xv=25 is a 
straight line. It is the line PQ. 
This line cuts the circle in the two 
points Pand QY. The codrdinates of 
the two points P and Q, where the 
line cuts the circle, are (—4, 3) and 
(3, 4). These are the two roots of 
the given equations. That they should be the roots 


appears from the fact that they are the only two points 
whose coordinates are the same for the line and the circle. 





II. Solve the following equations: 


(1) x + y? = 25. | 

(2) Sa+4y= 25. 

(3) a SAY, from (2). 
ioe 2 

(4) wee TN IO + y* = 25, by substituting in (1). 


(5) 625—200 y4+16 42+9 y? = 225. 
(6) 25 y? — 200 y + 400 = 0. 


22, THE ESSENTIALS OF ALGEBRA, 





(8) Y-HY-4)=9. 

(9) 3 y =4 and 4, 
(10) R20 a: hd ea oe 
3 3 
(11) x= 3 and 3. 


The roots are (3, 4) and (3, 4). 
‘These roots are the same. Equations (1) and (2) are 
sald to have a double root. 


The graphs of 224 y?=25 and 82+4y=25. 
a+ y%=25isthesamecircleasin(1). 32+4y= 26 is 
a straight line. ‘The graphs are 
shown in the adjacent figure. 
In this case the line PQ, which 
is the graph of 3872+4y= 25, 
p does not cut the circle, but just 
touches it at the point Y. The 
coordinates of the point @ are 
(3, 4). (8,4) is one of the two 
equal roots of the given equa- 
tions. In the case of equal roots the graph of the linear 
equation just touches the graph of the quadratic equation. 


&) 


III. Solve the following equations: 


(1) | w+ y2= 25. 
(2) x+y=10. 


(4) 100—20 y+y2+y2= 25. 


SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 233 


(5) 2y2— 2 y+75=0. 
20 + V400 — 600 
6 piesa Nee OY 
(6) y 4 
_ 20+10V—2 
4 
T10fhV =2 -1045V2 
Seana a Fe 
10+5V27 
(T) Ba a 
_10F5V27, 
2 


The roots are 


10—5V2¢ 104+5V2: J 104+5V2¢ 10—5V2: 
Mee Kg ae OLR 





These roots are both ¢maginary. 


The graphs of 27+ y?= 25 
and z+y=10 are shown in 
the adjacent figure. 
Inthis case the line PQ, which e 
isthe graphof w+ y=10, neither 
euts nor touches the circle. Q_ 
In the case of imaginary roots 
the graph of the linear equation 
neither cuts nor touches the 
graph of the quadratic. 


170. Graph of the Quadratic in x and y. From the pre- 
ceding discussion it must not be inferred that the graph 


234 . THE ESSENTIALS OF ALGEBRA. 


of the quadratic equation in x and y is always a circle, 
It may be any one of the following curves : 


OC )C FZ- 


Circle Ellipse Parabola Hyperbola 
Graph of 427249 y*= 36. 


From this 





When 2=0, y=2 and — 2; (0, 2), (0, —2) are roots. 
When z=1 and —1, pearey and — 4/2; 

CL4V2), (1, —4V 2), (= ey 2)nCea ~4V3) are roots. 
When z=2 and —2, y=8V5 and — aV5; 

(2, 2V5), (2, —2V5), (—2, V5), (—2, ~3V5) are roots. 
When a=3 and —3, y=0; (8, 0), (—3, 0) are roots. 


Locating these points and drawing a curve through 
them, we have an ellipse like the following. 


| 


SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 285 


EXERCISES. 


Solve the following sets of equations: 


e +y =5, (s%x—2y=8, 
a? + y? = 25. oil 4 wy = 32. 
meee = (eat2y=9, 
(2a +3y =6. eee ay 
3. eee . fa+3y'=27, 

aed ~ ( aty=10. 
4. ala of ay? = 26, 
ay = 50. 2. Ale y 296 
5. ae 
ay = 27. 13 ae: at Bae 
o—y" = 16; 
x —y =4, 
6. ; os 
ae sarubre 
Layty’=9 
, ee 
, ry == 0 e+2y=7, 
15.,3 10 
5 vw? — ay =2 


Construct graphs for Exercises 1, 2, 3, 5, 13. 


CASE II. 


171. Both Equations Quadratic of the Form ax’ + by’=c. 
When the equations are of this form, one of the variables 
may be eliminated as in simultaneous equations of the first 
degree, and the resulting equation is a pure quadratic in 
the other variable. 


236 THE ESSENTIALS OF ALGEBRA. 


I. Solve 
C1) rat yt see 
(2) 427+ 25 y*? = 100. 
Multiplying (1) by 4 and subtracting from (2), 


(3) 21 y = 36. 
(5) yotviz=+2vi. 
(6) 2+(+vi2)?=16, substituting in (1). 
(7) a= 16—-1= 1), 
19 
8 r= t—" 
*) V1 
The roots are (- 10 9 ve) Ge 24/8 =) 09 24/3) 
V7 v/ 
(= Semeny Vere 24/2 >) There are, as in all solutions under this 


case, a roots. 


Graphs of a+ y¥2=16 and 427+ 25y2=100. The 
graph of the first equa- 
tion is a circle and of the 
second an ellipse. ‘They 
are shown in the figure. 

The graphs of the two 
equations intersect in 
the four points P, Q, R, 
and S. The codrdinates 
of these four points are 
the four roots of the set 
of equations. 





SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 237 


I. Solve | Srey ay 

(2) 4a? +4 25 y? = 100. 
Eliminating z* as in Example I, 
(3) ge U, 
(4) y= 0. 
(5) y=0 and 0. 
(6) x? + 0? = 25. 
(7) e410. 


The roots are (5, 0), (5, 0), (— 5, 0), (— 5, 0). The 


set of equations has two pairs of double roots. 


Graphs of 2+ y?=25 and 4274+257?=100. The 
graphs are a circle and ellipse. They are shown in the 
adjacent figure. 

The graphs of the two equations do 
not intersect, but they touch each 
other at the points Q@ and P. The 
coordinates of these points are the 
roots of the equations. As in Case I, 
when the graphs touch, the codrdi- 
nates of the points where they touch 
are double roots. 


Chyiat 4? =I, 

(2) 422+ 25 y= 100. 
Eliminating 2? as in Example I, 

(3) Zyae 96: 

(4) y= Ft 

(5) y=tV3=24V4. 


les Solve 


238 THE ESSENTIALS OF ALGEBRA. 


(6) e+ (tVipt= 1 
v= 1-§t=— Hf 


The roots are (6V+¢, 4/2), (5Vti, —4Vv2), 
(—5V1i, 4V2), (—5V4t, —4-V2). 


These roots are all imaginary. 


Graphs of 2+ y2=1 and 427+ 25 y?=100 are a circle 
and an ellipse, respec- 
tively. They are shown 
in the figure. 

The graphs of the two 
equations neither inter- 
sect nor touch. The 

circle is entirely within the ellipse. As in Case I, the 
roots are imaginary. 


EXERCISES. 


Solve the following sets of equations : — 


a+ y?= 18, 5 {| v+37r=52, 
| (Qe + y=. ' (2a? 4+5y?=19, 
2 Pe hedg 
ieee ae 6 1 ee 
LC — a . 
e+ 2y?= 41. ae oe 
Ct ae 
3 (ety =%, BAA 
(3.024 44° = 180. ae ve 
| A 9 ein 
(raat ae ty ia esis 
+24? = 107. 3 Fi ae 


Construct graphs for Exercises 1, 6, 8. 


SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS, 239 


Case III. 


172. Both Equations Homogeneous in the Part involving the 
Variables. ‘he equations are of the form 
ax* + bery+ey2=hk. 
The first step of the solution is the elimination of the 
constant terms of the two equations. 





1) 222-—8 a 2, 
I. Solve , Me ee 
(2) 2a%-—3 y?= 838. 
Multiply (1) by 8, (2) by 2, and subtract. 
(3) 2a7—9ry+9y?=0. 
(4) Qar-—sy@—3y)=0. 
(5) ° ara Y. 
(6) B= FY: 
(7) w= By. 
Substituting z= 3 y in (2), 
BY 2 
(8) o(-# : i} res 8. 
2 
(9) ane ee 3 ye — oe 
By" 
(10) 5 ny 
(11) y" = 2. 
(12) y=+v2. 
(18) a= 8(+2)=+8V2 
Substituting 2=3 y in (2) 
18 472-38 y=3 
I= 5 
eos 
y= +}Vv5. 


a= 8(+1V5)= +3V5. 


240 THE ESSENTIALS OF ALGEBRA. 


The roots are 
(8V2, V2), (—8v2, —V2), GV5,4V5), (—2v5, —4FVb). 
In this case there will always be four roots. ‘There 
may be one or two pairs of double roots. Two or four of 


the roots may be imaginary. 
In this case the equations may be of the form 


ax* + bry + cy* = dx or ey. 
They are solved precisely as the above. 
(1) 2227-8ay+y7=2y, 
(2) 2279-38 y=3y. 


Eliminate the right-hand members by multiplymg (1) 
by 3, (2) by 2, and subtracting, 


II. Solve | 


(3) 2a7—-9ayt+9 y=. 
(4) (2e—By)\(@—3y)=0. 
(5) a=yand dy. 
Substituting z= 3 y in (2), 

(6) 2g yy? —- 8 y= By. 

T BY By, ya—-2y=0 2)=0 
(7) 9 8 Y 49 = 0 
(8) y= 0 and 2. 

(9) x= 0 and 8. 
Substituting z= 3 y in (2), 
(10) 9y—sy¥=3y. 
CLL) 6 y= 3 y. 

(12) y(2y—1)=9. 
(138) y = 0 and 4. 


(14) a= 0 and g. 


SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 241 


The roots are (0, 0), (0, 0)), (8, 2), (3, 4). (0, 0) is 
a double root. 

In each of these examples, after finding the value of 2 in 
terms of y, the substitution might have been made in the 
first equation instead of the second. ‘The second was 
selected because it was of simpler form than the first. 


EXERCISES. 
Solve the following sets of equations: 


ee - (2a°—ay+y7’?=16, 
eb = 5. a + ay +2 y? = 44. 


w+ 2 ey—y’=T, a Deed ie? 
22° —38ay—y =—1. y? + xy = 60. 
{sy—5x%=70, 
ly? — 3 ay = 10. 


la? + my’? =n, 


ev’ —y’? = 3, 10. , 
4. be ax + by? =. 
—2Zaeyt+2y=2. 

(4? +4y=—13 —4 ay, 


— a i Oe 
ca = 41, 8 2? —12 ay=11—8 77 


>= =37—y' 
sey 12 4247 12. : te 


aw 
2 
32° —9 xy —y’=119. (a 


+ = 73 — 2. 


CASE IV. 
173. When Both Equations are Symmetrical in x and y. 


Hquations are symmetrical in x and y when the interchange 
of these letters does not change the equations. 


242 THE ESSENTIALS OF ALGEBRA. 


Examples: (1) 2—-382y+y?=27T. Interchange 2 and y, 
and we have y? — 3 yx+ 2°= 27, which is the same as (1). 


(2) v—2ayt+y*=16, 

(3) a+ y* = 12, 

(4) i tLe 
and (5) xe+y=6, 


are all symmetrical equations. 


I. Solve | SAE TA a 
(2) 1A bie 
Add 2 zy = 24 to (1), and we have 
(3) + 2ay+y*%= 49, 
(4) a+y=x+T, from (38). 
Subtract 2 cy = 24 from (1), and we have 
(5) ew—2eyt+y=l, 
(6) e—y=H+1. 


From (4) and (6) we have, by adding and subtracting, 
z=4, 3, —4, —3, 
y =3, 4, — 3, —4. 
The roots are (4, 3), (8, 4), (— 4, — 3), (— 3, —4). 
When both equations are general quadratics, the solu- 


tion depends upon a cubic or quartic. The investigation 
of such equations is beyond the compass of this book. 


(CA) #f+2ey-—y/=T, 
12) e—2y>+y =2, 
Solving (2) for 2, 


(3) t= $V242 Py. 


II. Solve 


SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 2483 


Substituting in (1), 





(4) 242y—yt2yV2+2y—y—y=T. 
(5) t2yV24+2P—y=5—y ty. 
Squaring, 

(6) 8 y?2+8 yt—4 p= 254 y'—9 y?—2 yP +10 y. 
(7) Tyt—2y +17 Y—10 y= 25. 


This equation is a quartic in y, and unless it breaks up 
into factors of degree not higher than two it can not be 
solved by our present methods. 

Graphs of 27+ y¥?=25 and zy = 12. 

The graph of 27+ y? = 25 is a circle. 

ey = 12. 


When r= 0, y=; (0, &) is a root. 

When z=+1and —1, y=12 and — 12; 
C, 12), (—1, —12) are roots. 

When 2=+2 and — 2, y=6 and —6; 
(2, 6), (— 2, — 6) are roots. 

When z=+3 and —3, y=4 and —4; 
(3, 4), (-- 38, — 4) are roots. 

When «=4 and —4, y=3 and — 3; 
(4, 3), (— 4, — 3) are roots. 

When z=5 and —5, y= 22 and — 22; 
(5, 22), (— 5, — 22) are roots. 

When x=6 and —6, y=2 and — 2; 
(6, 2), (—6, — 2) are roots. 


244 THE ESSENTIALS OF ALGEBRA. 


When z=7 and —T, y=16 and — 12; 

(7, 18), (— T, — 18) are roots. 
When a= 8 and — 8, y=1) and —13; 

(8, 14), (— 8, —14) are roots. 
When «= 9 and — 9, y=14 and — 14; 

(9, 14), C(— 9, — 14) are roots. 
When a= 12 and —12, y=1 and —1; 

CL2t 1: (— 12, —1) are roots. 
When a2 = 24 and — 24, y=4 and —3; 

(24, 3), (— 24, — 4) are roots. 

The graph of zy = 12 
is the adjacent hyper- 
bola intersecting the 
graph of 22+ y?=5 in 
the points P, Q, AR, and 
iS, whose coordinates are 
the four roots of the 
equations, 

In — solving 
this example, 
we get the 
equations «+y=+T7 
and <—y=2 1 ie 
graph of x+y=7 is 
the line PQ, and that 
of «+y=-—T is the 
line RS. The graph of 
x—y=1 is the line QS, and that of z—y=—1 is the 
line PA. ‘These four lines intersect in the four points 










SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 245 


Pro h, and ‘S. 


The four lines have precisely the same 


intersections as the circle and hyperbola. ‘This is why the 


set 


is equivalent to the set 


as was used in the solution. 


2 y = + 1, 


a + y* = 25, 
ig) fend 8 


EXERCISES. 


Solve the following sets of equations: 


far + y= 13, 
ar = 6, 
a’ + y? = 34, 
xy = 16. 


xy = 24. 


| — 


al 

| habe 
9. | 

ing) ie 

| a y 

co 

ve Y 
lo. | 

roe ea 

[a y 

ae 5 
ae Ey 6 

Ei Sige aie 

Lae, eae 
zo [3 o+-5y = 2 ay, 
{ xy = 15 
“e Donna 
x+y =90 

er Yy=aA, 
14. 

fe = 1(a’?— 6’) 


Construct graphs for Exer- 
cises 4 and 7. 


246 THE ESSENTIALS OF ALGEBRA. 


174. Special Methods; Higher Degrees. Simultaneous 
equations of higher degree than the second can frequently 
be solved by special methods. ‘This is particularly true 
when they are symmetrical. 

A few of the special methods will be illustrated. In 
such problems the student is expected to devise his own 
methods. 


I. Solve 


(1)+(2) =) at + ay + oy? + cy + yf = 211. 
(2)4=(4) at —4a8y + 62%y?-— 4ay24 yt =1. 
(8)—(4), (5) 5aty — 52%? + Say? = 210. 
(6) a8y — ay? + vy? = 42. 
CT) xy(a? + y*) — ay? = 42. 
From (2), a —2ay + y2=1., 
e+ y=2ey+1. 
Substituting in (7), 
cy(2xy +1)— ay? = 42. 
ary? + xy —42=0. 
(cy + T) (zy — 6)=9. 
sy= 6 and <= (i 


6 ix 
t= = and =——s 
Y Y 


Substituting in (2), 


Daay arte and = a 
y y 

6 ay = ee 
Pt Yee yy ee 


(y—2)(y+3)=9, ae —-14+V-—27 
y=2and —3; 


SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 247 


may, 
The corresponding values of z are 3, — 2, Lea 


ye 
The roots are (8, 2), (— 2, — 3), 
ee Oe eas 87) ft nN OF 1 <1). 
7 0 er, are | F D 9 D) 
oo bees 
2 ee Aa 
II. Solve aes 
|@) peed a3: 
Oe 1 oat 
Lees be -9 
2) = (3) —— — + — = -— 
CYE@) a5 tia- 4 
‘5 
()-@)=(4) ==9 
ry 4 
1 2 eee 
1 i Dt = 


y CF, 
Combining (2) and (6) < addition, 


1 3 
—* and Dr 
z=4and — 2. 
y=% and —4. 
III. Solve - ne ees: 
Shaye yf == (. 
Put L=U+, 
Y=U—v 
Moe a Ad 
te 


at + yt = 2ut +12 u%?2420t= 641. 


248 THE ESSENTIALS OF ALGEBRA. 
Putting in the value of », 
2 ut + 12? (42) + 2(42)? = 641. 
2u4t + 147 wv? — 2727 = 0, 


4 
e 7 paar 
g=utv=d, 2, 5 + a 
T — 303 
y=u-—v=—2, —5, =p hae 
a+ y*= 10, 


IV. Solve es 
(2) vy—ax—y=-1. 


Multiply (2) by 2 and add to (1), 


(3) e+ 2ayt+y2?—-2e+y)=8. 
(4) (c+ yy? 2(e@ty)—8=0. 
(5) (e@+y-—)N@tyt+2)=0. 
(6) x+y=4or —2. 

From (2) by substituting the value of x +4 y, 
(7) ry = 9d, or — 3. 

Multiply (7) by 2 and subtract from (1), 
(8) | 2 —Qay +y2=A4, or 16. 

(9) a—y=+2, or +4. 


Combine (6) and (9), 


The roots are (8, 1), (1, 3), (1, —3), (—3, 1)- 


SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 249 


EXERCISES. 


Solve the following sets of equations: 


Pee 1 11 ees LS apg 
ah y * Lay =12. 
ee a ei Reb Mec 
: as? e—y=1, 
cy —y = 4, dal Carty = 82, 
= ees, | if “e+y=4, 
ied ss (a? +24? = BA, 
a ol ) | xy + y” = 35. 
a+ y=. ( 304+ 5y°=17, 
a? — y= 19, a Fs foe: ee Lp 
5. | es ame 
e—y=1 28 o 
a+ yi = 91, LeMah ean 
6. | ee 
e+y=7. 
aw + 4? = 28, 17. ie pres 
ae 2M Rage 
( xy + ay? = 30, sm J x 4-4? = 37, 
| aty=b. le+y+ay=13. 
pierre ay, he 
x—y=1. 
— 2 —s = 
10 { MN +3 @—N=18, 20 | ae 35, 
a+ y =T. (@+y)(@ +y") = 65. 
EXERCISES. 


1. Find two numbers whose difference is 5 and the differ- 
ence of whose squares is 145. 

2. The difference of two numbers multiplied by the greater 

= 100, but multiphed by the less = 84. Find the numbers. 


250 THE ESSENTIALS OF ALGEBRA. 


3. The sum of two numbers is 7, and the sum of their cubes 
is 91. Find the numbers. 


4, The product of the sum and difference of two numbers 
is 96, and the sum of their squares is 146. Find the numbers. 


5. The sum of two numbers multiplied by their product is 
120; and their difference multiplied by their product is 30. 
Find the numbers. 


6. The difference of two numbers is 3, and the difference of 
their cubes is 117. Find the numbers. 


7. The sum of the areas of two square fields is 2500 square 
rods; the sides of the fields are to each other as 5 to 4. Find 
the area of each field. 


8. If the length and width of a rectangular field are each 
increased 10 rods, the area is increased 5 acres. But if the 
dimensions are each decreased 10 rods, the area will be 24 
acres. Find the dimensions of the field. 


9. The diagonal of a rectangle is 180 feet; the length 
of the rectangle is 22 times the width. Find the dimen- 
sions of the rectangle. 


10. Find two numbers such that their product is 16 times 
their difference, and one of the numbers is double the other. 


11. A rectangular lot containing 13200 square feet is sur- — 
rounded by a walk 6 feet wide. The walk contains 3336 
square feet. Find the dimensions of the lot. 


12. In acertain number of two digits the sum of the squares 
of the digits is one more than twice their product, and the dif- 
ference of the squares of the digits is 7. Find the number. 


13. The fore wheel of a carriage makes 12 revolutions more 
than the hind wheel in going 240 yards; but if the circumfer- 
ence of each wheel is increased 1 yard, then the fore wheel will 
make only 8 revolutions more than the hind wheel in the 
same distance. Find the circumference of each wheel. 


SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 251 


14. If a man had worked 5 days less and had received $1 
a day less, he would have earned $30. If he had worked 10 
days less, and had received $2 a day more, he would have 
earned $50. How many days did he work, and what were 
his wages a day ? 


15. Ifthe numerator of a fraction be increased by 3 and the 
denominator be decreased by 3, the resulting fraction is the 
reciprocal of the first. If 47 be added to the fraction, the sum 
is } the reciprocal of the fraction. Find the fraction. 


EXERCISES— MISCELLANEOUS. 
1. Extract the square root of 2’y? — ary? —(4a—Ly)a’y 
+2ay+4a' 
2. Find the roots by factoring: 
(a) # —Tx=390. 
(6) #+72=60. 
(c) y—9ay+20a=0. 
(d) By+4(2y—3)—39=0. 
(e) y—(e—a(e—b) = (a—D)y. 
3. Determine whether 1, —1, 2, or any one of them, is a 
root of 9a’°—3a=2. 


4. Make an equation whose roots are 3+ -V7 and 3— v7. 
5. Simplify 8-V3 + 13-243 — 5-121 +.4-V27. 

6. Multiply (@+Va7) by a? —Va~. 

seeLE SV Dia. 

2V5 —3-V6 
34+V—4 
6—V—16 


7, Rationalize the denominator of 


8. Rationalize the denominator of 


1 
3 


9. Divide a—b by a’ — 3. 


Oo THE ESSENTIALS OF ALGEBRA. 


Seon 
v42%=8, 
vo sy 

10. Solve ee 
Ge oe 4, 
Ey, 


11. Solve de?—2x+c=0. 

12. By means of the discriminant, tell what kinds of roots 
each of the following equations has: 

(a) 3a°—5a+2=0. (c) 5a°—5a+10=0 

(b) 22?+112—10=0. (d) —a’+4e+4+2=0. 

13. Find two consecutive numbers whose product is 1260. 


14. A number consisting of two digits which differ by 3, is 
6 less than 7 times the sum of the digits. Find the number. 


15. What value must a have to make the roots of 52?—112 
+a=0 equal ? 
2 ey, 


——-=0, 
a1 fd eed Find « and y. 
9 
+55} 
“Ber ir: Find « and y. 
dy—v7+3=0. 


18. Two trains start at the same time to go 320 miles. One 
goes 8 miles an hour faster than the other and reaches its 
destination 2 hours sooner than the other. Find the rate of 
each train. 
e+2 “+3 


eo 1 
=p oy 
PINE at eel 











19. Solve 


20. Express 27 ys 4 2aiy7# without negative or fractional 
exponents. 


21. 120*—172°4+6=0. Find z. 
22. y°—3d ye == O6..q Ing: 


SIMULTANEOUS EQUATIONS INVOLVING QUADRATICS. 253 


23. A number consists of two digits. If the order of the 
digits be changed, the sum of the new and original number is 
77 and their product is 1300. Find the number. 


24. 36074 29 ax+5a°?=0. Find z. 
25. Make an equation whose roots are 
26. V2 +27 =8. Find 2. 
erie oO ee 0 —C. 
b+ta w-—a b+c we-—Cc 
28. Find the roots of (y—2)(y? —12 y + 20)(y—1) =0. 

29. DV — 2? x 8V— 7 x 2V—9 ce? x 8 — 48? = what? 

30. Multiply V3 —V2—vVy by V3+ Va— Vy. 

31. Perform the indicated operations and simplify the 


result: 
Ve= (aa 
a Gee 


5 


32. Square ai + bt — 3, 
33. Solve 11 x—11= ae 


a+v2 and a—Vv2 
3 3 


27; Find a. 

















34. Solve 22—352+4+22°=0. 
a ws 5. 2a+3 
—5 2a—-3 
a+5 a= ba 
2a+3 Seen 
36. A path around the outside of a rectangular garden is 6 
feet wide and 4224 square feet in area. The area of the gar- 
den is 28000 square feet. Find the dimensions of the garden. 








35. Simplify — 








Byipsinplity 22. 


954 THE ESSENTIALS OF ALGEBRA. 








A Bs Ob 
aa Solve eee A 
Ohh Se OR es ed 6 
3 2 ace tie der 2 a2 — 14 
39. vipers are Nae ’ Find @# and y. 


2y—dSay+ouv= 6. 

—6 2 y—1 2 
40. Solve &— a : 
Abr AD RMON pe aE So 


41. Solve (a? — b’)(a—1) =2 a(a’ + 0”). 


42. Solve s]2—4{2—a(2—- 2) | | as. 


43. What must be the value of x in order that 
may equal —1? 

44. Find the value of at—5a*—122?—13%—T7, when 
g=—i(14+~V—8). 

45. Two rectangular fields each contain 10 acres. The 


perimeter of one is + longer than that of the other. One of the 
fields is asquare. What are the dimensions of each field ? 














(w+ 3)? 
32°+9a—5 


46. If ab+bce+ca=0, prove that 
(a) (a+b+cPr=e?40?+e%. 
(6) (a+b+cPf=e04+?+¢—3 abe. 
(c) (@+b+c)=ai+b*+ct—4abe(a+b+0). 


CHAPTER XVII. 


RATIO, VARIATION, AND PROPORTION, 
I RATIO. 


175. The ratio of a quantity A to a quantity B is the 
quotient of A by B. 


This quotient may be written in any one of the forms, 
A + B, 4 A/B, or A:B, each of which is read, the 
ratio A to B. 

176. Ratio can exist only between two abstract num- 
bers, or between two concrete numbers of the same kind. 
The ratio 5 to 7, or ? , has a meaning, so does the ratio 
6 bushels to 15 bushels, but not so with 6 bushels to 15 


inches. Ratio merely expresses the part one magnitude is 
of another. 


177. The terms of a ratio are the numbers compared, 
the numerator being called the antecedent, the denominator 
the consequent. 


178. The ratio of antecedent to consequent is called a 
direct ratio; the ratio of consequent to antecedent is called 
an inverse ratio. 

Thus, 14: 28 is direct, while 28:14 is its inverse ratio. 


b is the inverse of ©. 
a b 


255 


250 THE ESSENTIALS OF ALGEBRA. 


179. A compound ratio is the product of two or more 
single ratios. 


Thus, 5 x is the compound ratio of the single 


bh gh 
<a *f 
ratios, ~ “ # 

5 b’ Teck 


180. Laws of Ratios. Since a ratio is a fraction, the 
operations which may be performed upon fractions may 
likewise be performed upon ratios. Below are enumerated 
the more‘important laws relating to ratios: 

(1) A ratio is unchanged by multiplying or dividing 
both antecedent and consequent by the same number. 


MeO oe 00-2 ee mA A-+-m 
§ 8x4 849° Bo mB Boome 











This law shows that the ratio of two concrete magni- 
tudes of the same denomination is independent of the unit 


of measurement. ‘The ratio of 2 miles to 5 miles is 53 the 


ratio of 2 miles, expressed in feet, to 5 miles, expressed in 
2x 5280 2 
5 x 5280. 5 
numerator and denominator of a ratio merely changes the 
denomination of the terms, if they be considered as con- 
crete quantity, the ratio of the two magnitudes remaining 
unchanged. 

(2) A ratio is changed by extracting the same root of 
each term of the ratio, or by raising each to the same 
power. 

This law is true except in the case when antecedent and 


feet, is 





The introduction of a multipher in 


A VA 
consequent are equal. If pa then will TR = /m, and 


RATIO, VARIATION, AND PROPORTION. 257 


k a 
se =m*; but Vm#m, and m‘*#m. Hence, the ratio has 
been changed in all cases except that in which m= 1. 

(3) A ratio is changed by performing unlike operations 
upon antecedent and consequent. 


ees es 


— + ———, in which the antecedent has been multipled 


AE re 
A mA mA (m\A 
by 2 and the consequent by 3. —+——, for ——= (“) at 
sf : 7 Bb 7 aB nB 1} BD 
(4) A ratio is changed by adding the same quantity to 
antecedent and consequent, except when the ratio is unity. 


BoA BALL oe 


7 ORI apr ey igs 





The truth of this law may be easily proven. Take the 
A+” 
B+a 





: A 
two fractions, 2 and , where x is any number what- 


ever. By division, 
A+z A, x(B—A) 


B+« B BCB+2) 
Alte 
Brow 
(5) A ratio is made more nearly equal to unity by 
adding any positive number to each of its terms. 








which shows that oe except when A = B. 


Let 4 be any ratio, and z any positive number. Then 








A+a _ ve ee 

B+o2@ Bia 

and Ay why eres 
B B 


The first of these two differences is seen to be smaller 
than the second. Why? Hence, the truth of the law 
is established. 


ih) THE ESSENTIALS OF ALGEBRA. 





(1) Compare = and 











2_20 243 5 _ 86, 
TALT0 SU oe LOM 
2 
= is more nearly 1 than = 
(2) Compare > and ote. 
DUS BOU Di toes en 
8 21° 344 7 21 
30 


ae 
nA is more nearly 1 than a 


os 


From these illustrations we may see that if a ratio be 
less than unity, the addition of the same positive number 
to the antecedent and consequent increases its value toward 
unity; and if the ratio be greater than unity, the addi- 
tion of the same positive quantity to both antecedent and 
consequent diminishes the ratio toward unity. 


181. The terms ratio of less inequality, ratio of equality, 
and ratio of greater inequality are sometimes employed to 
describe ratios less than unity, ratios equal to unity, and 
ratios greater than unity, respectively. 


182. Limit. The result shown above, 


wales a clas ps Par ee oe) 
B4+z Bax 








A+z 





indicates that the difference between and unity 


is a fraction 





whose value may be made as small 


as we please by making @ sufficiently large. Hence, the 


RATIO, VARIATION, AND PROPORTION. 259 


A+z 


value of the ratio -, as x becomes infinitely large, 
x 





approaches unity, which is called the lamit of the ratio. 
The value that any algebraic expression continually 
approaches -but never reaches is called its limit. 


EXERCISES. 


Write in their simplest forms the ratios of: 





1. 625 to 125. 4. 2 —(y+z2) to ~+y+z. 

2. 4802 to 120 2”, 5. 2—y to a+tay+y’. 

3. 2432? toxr+3. 6. (‘St 1) A a 
4 ab 

SUGGESTION. pee (et 8) x, 


r+3 r+3 
7. a@—12a+ 20 to a—10. 
8 627+ 23 ax +20 a? to 32+4a4. 
9. wt +ary’?+ y* to v—ayt+y’. 
10. (a+ 7’)? —4 a°y* to (a? — y’)’. 
Write the compound ratios of the ratios: 
11.35 to 5 and 10 to 15. 
12. ~+y to «—y and w—y7’ to (#+ ¥)*. 
13. 25 to «* and 32’ to 50. 
14. a? — 270° to (a—35b)? and a—3b to &+3ab4+90". 
15. (©+1)?: @’+2a+41), (#’+1):(#+1), 


and (a—1) : (#’—a+1). 
Find the value of x for which the ratio of: 
16. 128 to 2 is 2. 1602+) 60,4 — 1 18.7. 
Te O20 LO 2 18 0. 19500 + 1207) to: 27-Eb- is. 3. 


20. (7+ 4):(8e%41)=4. 


260 THE ESSENTIALS OF ALGEBRA. 


Arrange the following ratios in descending order of magni- 
tude : 





21 9 21 AS er I Yeah EF 8 4 1 
© 2:02 PS22 9 '2 GF. 92 Fee D4) oO iy 2 eee 
22 ats “a+ @-+3) aie 

















II]. VARIATION. 


183. The term variation has little use in ordinary alge- 
bra, but its use is so frequent in physics that a brief treat- 
ment of the subject will be introduced here. 

In physics we say “the weight of a uniform mass varies 
as the volume.” ‘This means that if W is the weight, and 
V the volume, then is W=k x V, where k is a constant, the 
weight of a unit volume of any given substance. 

In mensuration the circumference varies as the diameter. 
This means, that if ( be the circumference and D the 
diameter of any circle, then will 


C=kx D, 
k being a fixed constant for all circles. ‘This constant is 


usually denoted by the Greek letter 7; its numerical value 
is an incommensurable number, 3.14159 «--ccccceeeeeseee : 


184. In general a variable y is said to vary as another 
variable 2, when 
“= a constant. 
x 


The phrase, y varies as x, is sometimes written 
Yer, 

but is to be interpreted to mean 
wi 


x 


i NOT eae, 


From this definition we see that a variation as here con- 
sidered is equivalent to an equation. 


RATIO, VARIATION, AND PROPORTION. 261 


185. Variations may be classified as follows: 
(1) Direct. y varies directly as x, when 
y=kx, k=a constant. 


The circumference of a circle varies directly as the 


radius. 
C=kR, where k= 27. 


(2) Inverse. y varies inversely as x, when 


y=: 
es 
The volume of a gas varies inversely as the pressure, 


k 
V=—, where V= volume and n= pressure. 
n 


(5) Joint. y varies jointly with xv and z, when 
Dire abe 
The weight of a rectangular parallelopiped of metal of 
unit height varies as the product of the length by the 


width, 
Veer delle xan). 


k = weight of unit volume of the substance, / = length, and 
6 = width of the rectangular solid. 


(4) Quadratic. y varies as the square of 2 when 
URW Le 


An example of such variation is found in the law of 
falling bodies; 7.e., the space fallen through by any body 
starting from rest equals a constant times the square of 
the time expressed in seconds. 


w= ki?, k=tg9, g = 32 feet, 2 inches. 


S'= space described, t = time in seconds. 


262 THE ESSENTIALS OF ALGEBRA. 


(5) Direct and Inverse. y varies directly as x and in- 

versely as 2 when | 

= k-. 

Yer 

An example of this form of variation is found in 

Newton’s Law of gravitation. If M, m, be the masses of 

two attracting bodies, D their distance apart, and G the 
force of gravitation, then 


Mx m 
G = k——— 
Dp 
EXERCISES. 


1. If yoo, and y=b when «=a, find the value of y when 


2 pint 
SOLUTION. 


If yea, thenisy=kx. Buty =b when x=a; 
b 
bence, b=ka, or k=-- 
a 
a 2 for any value of zx. 


b 
Hence, y=—-c when x= c. 
a 


2. If you, and if y=10 when x=2, find the value of 9 
when «=12. 

3. If you, and if «=16 when y= 64, find the value of a 
when y= 15. 


4. The circumference of a circle varies as the radius 
(Cx). If C=3.1416 when R=4, find the circumference 
of a circle whose radius is 12. 

5. If excy and wz, prove ~ «4. 

Mies 

6. If xxy and vet, prove avcyt. 


RATIO. VARIATION, AND PROPORTION. 263 


7. If xw«y, prove that zy”. | 
8. If wxy and zcy, prove that (a — 2°) oy’. 


9. If y varies inversely as a’, and if y=16 when «= 4, find 
2 when y= 10. 


10. The volume of a sphere varies as the cube of its radius. 
If the volume of a sphere whose radius is 3 be 113.1, find 
the volume of a sphere whose radius is 20. 


Ill. PROPORTION. 


186. Proportion. The equality of two ratios is called 
a proportion. 


Thus, a=< is a proportion. 


Various forms have been employed in writing a propor- 
tion, the following being the ones more frequently used: 


foe C.D), are Balad. 
i eee: 
A+B=C+D, eat 


Each form is read A is to Bas Cis to D. 


187. Proportionals. The terms of the two ratios are 
called proportionals. 

In the proportion A: B=C: D, the terms A and D are 
called extremes, the terms Band Care called means, of the 
proportion. 

In the proportion A: B= B: D, B is called the mean 
_ proportional to A and D; D is called a third proportional 
to A and B. 


188. Theory of Proportion; Theorems. A Theorem is a 
statement of a truth to be proved. 
A Corollary is a truth derived from the proof of a theorem. 


264 THE ESSENTIALS OF ALGEBRA. 


The following theorems apply to proportions in which 
the terms of each ratio are considered abstract numbers. 


THEOREM I. In any proportion the product of the ex- 
tremes equals the product of the means. 


Given A: B=C: D, or, more simply, 
AEC. 
Bed: 
hens A= Bes WW hye 


CoROLLARY. The mean proportional to two numbers 
equals the square root of their product. 


This corollary results from the above theorem by letting 


C= 8, a=, or AD= B*, whence B=VAD. 


THEOREM II. Jf the product of two numbers equals the 
product of two other numbers, then either pair may be taken 
as extremes, and the other pair as means, of a proportion. 


(Inverse of Theorem 1.) 
Given AD=BC. 


Divideshy 7 ale 4-5 
Divide by CD, (2) fas. 
Divide by AG, (3) 4 = 7m 


In each of the proportions (1), (2), (8), we have taken 
one pair of factors, A, D, or B, C, as extremes, the other 
as means. 


——— eS eee 


RATIO, VARIATION, AND PROPORTION. 265 


THEOREM III... Jf four numbers be in proportion, they 
are in proportion by inversion. 


Expressing this theorem algebraically, 


D 


= = oe from which B =i. 


tee D Ae 


Proof is left to the student. Result is easily shown ° 
true from Theorem I. 


THEOREM IV. Jf four numbers be in proportion, they 
will be in proportion by alternation ; that is, the first ts to 
the third as the second is to the fourth. | 

Pigepraically, if A: B=C:D, then.is A:C=B: D. 

See (2) under Theorem II. 


THEOREM V. Jf four numbers are in proportion, they 
are in proportion when taken by composition; that is, the 
sum of the first and second is to the second as the sum 
of the third and fourth is to the fourth. 


This theorem stated algebraically is, 
A Sault Ee a Ee 


if — = fa then is 


B D B D 
Proof is easily derived by adding 1 to each member 
of the given proportion and reducing each member to a 
fractional form. 








THEOREM VI. In a series of equal ratios the sum of 
all the antecedents is to the sum of all the consequents as 
any antecedent is to its consequent. 


Proof. Let the equal ratios be 
— Ss ea as - =" #** =7, 


where ¢ is the common value of the ratios. 


266 THE ESSENTIALS OF ALGEBRA. 


Then oar, or A=r xB, 
f=, or C= nxil; 
Ser, or H=r KF, 
amr, OF = cae 


Adding the equalities, 





A+C+H+G+--=7x(B+D+F+7aR 
ry A+C+H#H+G+4+>. AL 

H 9 —-- > = —  — See, 
BE D+F tps 1p 


I Eee aK! 


then is 
1 §+14+35+44+6 8+4 54146 


—_ rd 


3° 1643494129518 9419" [oe 


and so on. 





EXERCISES. 

Find the value of the variable for which each of the following 
proportions is true: 

LY O20 95-40: 
ee Dt 0. 520s 
edd = D2 a. 
(8a+4):(@+5)=(5e%+1): Ga—4). 
(4e2—3):(2e+1)=(7 7-4): (844 2). 


6. Find a fourth proportional to 12, 16, and 40; also to a, 
b, and c. 


Ses 


RATIO, VARIATION, AND PROPORTION. 267 


7. Find a mean proportional to 16 and 49; also tol and m. 
8. Find a third proportional to 25 and 35; also to a and «wz. 


Are the following proportions true for all. values of the 
letters: 


9. 9—2):(842)=(8x—2"): 4? 

2 
10. (Fat): 2ae—N =a): dee? 
ll. (w+ y)?—2): (e@+y+2)=(@+y—2): a? 
etme « 


Sa yaa 2 + .w, show a= 
lyt+mw y 





et Zz 
suGGESTION, Let —~=7r, —=r, then x=ry, z=rw, also v= Iry, 
v 


a 
mz=mrw. Add, lx + mz=r(my + mz), ete. 


2 2 1 re gl A 
13. If -=—, show ———.~=—=-—. 
w yw. yw. yy 
14. The rates of walking of two travelers are to each other 
asatob. If one walk c miles in a given time, how far does the 
other walk in the same time ? 


15. The rear wheel of a wagon is a feet in circumference, 
the fore wheel is } feet in circumference. How often does the 
fore wheel rotate while the rear wheel makes m revolutions ? 


CHAPTER XVIII. 
PERMUTATIONS AND COMBINATIONS. 
I. PERMUTATIONS. 


189. This subject can best be understood by introduc- 
tion through a few concrete examples. 

(1) How many different numbers of two digits each 
can be formed by using in every way any two of the five 
digits %, 0, seo.) ore 

By writing any one digit first and each of the remaining 
four digits after it, we have the following five rows, each 
composed of four numbers: 


56, 57, 58, 59, 
65, 67, 68, 69, 
195, 76, Tek 19; 
85, 86, oh SU. 
95, 96, 91, 98. 


In all there are 5 x 4= 20 different numbers. 

(2) How many different numbers of two digits each 
can be formed by using in every way any two of the four 
digits 0, 0.) 7,10 ¢ 

Here we select any one of the four digits as the first, 
and place after it successively every one of the remaining 
three digits. 

268 


PERMUTATIONS AND COMBINATIONS. 269 


- This gives the following numbers: 


56, 57, 58, 
65, 67, 68, 
795, 76, 78, 
85, 86, 87. 


In all there are four selections of the first digit, and four 
less one selections of the second, giving 4x 38=12 different 
numbers. 

(3) How many different numbers of three digits each 
can be formed by using in every way any three of the five 
PIPING t 4 Oy OD? 

The first digit can be any one of the five; hence, the 
first place of each number can be filled in five different 
ways. Four digits remain to fill the other two places of 
each number. But we have just seen that two digits can 
be selected from four in 4 x 3=12 ways. Hence, with 
each of the five selections of the first digit, can be placed 
twelve selections of the digits filling the two remaining 
places. Hence, there are 5x4x3=60 different numbers. 


190. Definitions. (1) The number of ways of selecting 
three things from a collection of five things is called the 
permutations of five things taken three at a time. 


(2) The number of ways of selecting r objects from a 
collection of n distinct objects, regard being had for the order 
of selection, is called the permutations of n things taken r 
at a time. 


In this general case, may be any number of objects, 
and r may be any integral number from 1 to n. 


270 THE ESSENTIALS OF ALGEBRA. 


191. Symbol. Instead of writing the permutations of 
n things taken r at a time, the symbol ,P, is generally 
used. 


Illustrations. (1) ,P,=5 x 4, the permutations of five 
things taken two at a time. | 

(2) ,P, =4 x 3, the permutations of four things taken 
two at a time. 

(3) ,P3,=10x9x8, the permutations of ten things 
taken three at a time. 


192. Hzamples: Let the pupil construct tables, if 
necessary, to verify the following results : 


Clit ley eens 

Aig Pes: 

(3) Fe Baa eg Ame ce 

(4)° Pe xt aes 

COs) Aiea. 

CDi): a Fg aha 

(Ty Pep es) 

(8) Pp AX oer 

(9)\ SP as 5x4 Xo eee 

CLOT Rebs Lue 
The above examples indicate that there is a law govern- 

ing the formation of permutations. It will be noted that 
the number of factors giving the permutations in each 
case equals the number of objects in each selection ; the 


highest factor is the number to be permuted, and each 
succeeding factor is one less than the preceding. This 


PERMUTATIONS AND COMBINATIONS, Atel | 


observation should lead one to some conclusion regarding 
the value of the general symbol 


atone 
We should expect the number of permutations of » things 


taken 7 at a time to be expressed by a product of 7 of the 
natural numbers beginning with n. Hence, we should find 


A Oe 

(2p aoe n(n — 1). 

(8) ,P3,=n(n—1)(m— 2). 

(4) ,P,=n(rn—1)(n—-2)™m-—8). 

(5) ,P,=n(n— 1)(n— 2)(n— 3) (m— 4). 


(r) ,P,=n(n—1)(n—2)(r— 3) « (n—r+1). 


193. Value of ,P,. ‘I’o determine the number of permu- 
tations of n things taken 7 at a time, we may proceed as 
follows: 


(1) Let the n distinct things be represented by n letters 
of the alphabet. 

(2) Select any one letter to stand first in a set of words 
of two.letters each. Then there would remain n — 1 letters 
to fill the second place; but the first letter may be selected 
in nm ways, and with each of these selections any one of 
the n — 1 remaining letters may be placed. 

(3) Hence, for the number of permutations of n things 
taken 2 at a time we have 


anal): 


(4) Let the first two letters of a three-lettered word be 
selected from the n letters; this selection can be made in 


O12 THE ESSENTIALS OF ALGEBRA. 


n(n —1) ways, as shown in (3) above. Now we may 
select any one of the remaining n— 2 letters to fill the 
third place. 

(5) Hence, the formation of a three-lettered word from 
n letters can be accomplished in 


nes =n(n—1)(m— 2) ways. 

(6) In a similar manner we may show that 

ne ,=nPs x (n—3)=n(n—1)(n—2)m— 8), 

»P,=nP, Xx n—4)=n(n—1)(n— 2)(n— 8) (n— 4), 

and in general 
per pee Sheet aD 
=n(n—1)(n—2)(n— 38) + (m—r+1). 

This general result may not be understood at the first 


reading of this subject, but its truth may be assumed until 
the pupil has had more experience. | 


CoroLuARY I. Whenr =n, the general formula becomes 
nen =n(n—1)(n—2)(m—3)4x38x2xI, 


a result easily remembered. 


194. The Factorial Symbol. In the value of ,P,, above, 
we have the product of the natural numbers from 1 to n. 
This product is often spoken of as “factorial n,” and is 


written for brevity 
[n, or nl. 


Either |n or n! is to be read factorial n, and means 
the product 


n(n —1)(n—2)(n—38)+5x4x38x2x1. 


PERMUTATIONS AND COMBINATIONS. ie 


| EXERCISES. 
Find values for: 




















sae eee 16 Ae 
[5 5, [ie 8. [25 e 
[6 [8 [8 Bale 
2. B | 
| 10 6 as 9 ee 
Spel abe 115 |4 oh 4s 
fate [15 [4 i 
20 20 8 {10 
4. Eu zs ue 10. es 








[16 x [2 [10 [12 [15 


195. Coro.tLARY II. When objects are permuted all 
together, but are not all different, the number of distinct 
permutations is given by ,P,,+|s, where s is the number 
of objects which are alike. 


Illustration. Required the number of different numbers 
obtainable from the five digits 5, 6, 6, 7, 8, taking five 
at a time. 

If all digits be different, the number of selections would 
clearly be ,P;=|5. But the two sizes, when permuted, 
give no new numbers; hence, all the permutations of the 
two sixes, z.e. |2, must be excluded (divided out) from 
the total. 





Pepe 2 Xt x8 x12 5 
Sa Peis 
EXERCISES. 


1. How many different numbers of three digits can be made 
from 1, 2, 3, 4, 5, 6? 

2. How many different permutations can be made by taking 
4 of the letters of the word working ? By taking all of them ? 


274 THE ESSENTIALS OF ALGEBRA. 


3. Find the value of i.P33 1;£s3 »Ls 

4. How many permutations can be made from the 26 letters 
of the alphabet, taking + at a time ? 

5. How many six-place numbers can be formed from the 
Arabic numerals? (Include 0.) 

6. In how many ways can a class of 6 be seated in a row of 
6 chairs ? 

7. In how many ways can the front row of 6 chairs be 
filled from a class of 20? 

8. How many different permutations can be made from the 
lette1s of the word Indiana? Mississippi ? 

9. How many even numbers of 6 places can be formed from 
the digits 1, 3, 4, 5, 7, and 9? 

10. How many numbers between 50,000 and 60,000 can be 
formed from the digits 3, 4, 5, 6, 7 ? 

11. In how many ways can 10 books be arranged on a shelf 
provided 2 particular books are always to be at the ends of 
the shelf ? 

12. In how many ways can 12 balls be arranged, if 5 are 
red, 4 white, and 5 blue ? 


Il. COMBINATIONS. 


196. (1) How many products of two factors each can 
be made from the five digits 5, 6, 7, 8, 9? 

We have seen that the number of ways of selecting two 
things out of five is the permutations of five things taken 
twoatatime. But in the case of products, 5x 6=6 x 5; 
hence, each arrangement of two digits is the result of a 
permutation of two things taken two at a time. These 
must all be excluded. Hence the number of products is 


5 x4+|(2=5x4+2=10. 


PERMUTATIONS AND COMBINATIONS. Bes 


(2) How many products of three factors each can be 
made from the five digits 5, 6, 7, 8, 9? 

Since any three factors may be arranged in 3 x 2 x 1 
different ways, each arrangement giving the same product, 
we shall have to divide out |8 of the permutations of 
five things taken three ata time. Hence, the total number 
of different products is 


ox4x3+|3= 10. 


197. Definition. (1) The number of ways of selecting 
three things from a group of five, no regard being had for 
‘he order of selection, is called the combinations of five things 
saken three at a time. 


(2) In general, the number of ways of selecting r things 
from a group of n things, no regard being had for the order 
of selection, is called the combinations of n things taken r 
at a time. 


198. Symbol. Instead of the phrase, combination of n 
things taken r at a time, the symbol ,C, is usually em- 
ployed. 

Thus, ,C, is read, the combinations of five things taken 
two ata time; ,,(, is read, the combinations of ten things 
taken four at a time, ete. 


199. Relation between ,C, and ,P,. It is easy to see that 
if we select from a given number of things any specified 
number, and do this in every possible way, having no 
regard to the order of selection, and then permute all the 
objects in each group in every way, we shall have the total 
permutations of the n things taken r at atime. ‘The selec- 


276 THE ESSENTIALS OF ALGEBRA. 


tions of the groups are the combinations, and the objects 
of each group are permuted r at a time; hence, 
os 4 
AY edema b MEN ay EPe 
OMY ROM 
MW aU. x 7. 


er, al, n(n) n= 2) Gree 
n~7 |” |” 


A second form for ,C,, may be had by multiplying the 
numerator and denominator of the fraction on the right 

















by |jn—vr. This multiplier makes the numerator |n, and 
the symbol ,C;, becomes 
cee 
“" |r |n— 7 
12 |12 
Cc = = e 
aie Khe ei ee 
From the first form 
| yy _12-11-10-9.8-7-6 
1277 [7 


12-11 10-9. Sy aeeeee 





Notr. When r=n, the second factor of the denominator |n — 7 
becomes |0, a symbol whose value is to be taken as unity. 


To show 
jo =1, 
we take the equality [m =m x |m —1, 
and put = 


PERMUTATIONS AND COMBINATIONS. pag 


then (has deta; 
and as [1 = 1, |O must be 1. 
oe |O =~ i 
The fo in Lie ee 
[2 [therm 


is mor? easi 7 remembered than the form 


C, =m = UG — 2)ur ea —r +1) 
a 9 


but the latter is especially useful in many applications. 





n 


EXERCISES. 


1. How many combinations can be made from 9 things 3 
atatime? 5 ata time? 

2. Find the values of Cy, Co, 12C3. 

-3. In a meeting of 20 people, in how many ways can a 
committee of 5 be selected ? 
4, A school is composed of 19 boys and 25 girls. In how 
many ways can a committee consisting of 1 boy and i girl 
be selected ? 

5. From the above school, how many comm.ttees consisting 
of 2 boys and 1 girl can be selected ? 

6. From 15 persons, how many committees of 5 can be 
formed, provided one particular person is to be a member of 
every committee ? 

7. If out of 9 candidates there are to be 5 officers elected, 
how many different tickets can be formed ? 

8. From 4 vowels and 8 consonants, in how many ways can 
5 letters be chosen, provided exactly 2 of them are vowels ? 
Provided at least 2 of them are vowels ? 

9. How many even numbers of 4 places can be formed from 
the digits 1, 2, 3, 4, 5, 6, 7, 8? 

Recs. C1; find n. 


CHAPTER XIX. 
SERIES. 


200. General Definitions. (1) Any set of numbers is an 
array, Or Succession. 

(2) A series is a succession of numbers arranged ac- 
cording to some law. 


Thus, 1, 2, 3, 4, 5, 6, ---, is a series, the law of formation 
being that any number is to be had from the preceding 
by adding 1. 

We may also define a sertes as a succession of numbers, 
the knowledge of two or more successive ones being suffi- 
cient to determine all. 

Thus, 5, 7, 9, 11, ---, form a series, since by inspection 
of any two we see their difference to be 2; hence, any 
number of the series may be had from the preceding by 
adding 2. 


(8) The numbers forming a series are called the terms 
of the series. 
EXERCISES. 
What law of formation exists in each of the following? 
1, 2, 4, 6, 8, 10, ---. 3. 4, 1, §, 2, §,.3, £, See 


2. 5, 10, 20, 40, 80, ---. 4. 3, 3, 2, 3, Be, +. 
278 


 — ae 


th i ti 


SERIES. 279 
Bani 1 a, Oy 1, Os 22% 
6. a, a+d, a+2d, a+3d, »-, a+ (n—1)d. 
7. a, ar, ar*, ar, «+, ar", 
gs. 5, —15, 45, —135, 405, — 1215, 
y oe x ag a 


Ben ey (8, 


nig ta tat 


ee 8. 


(4) When the number of terms of a series ts finite, the 
serves is called a finite series. 


Thus, 2, 5, 8, 11, 14, is a finite series. 
(5) When the number of terms is infinitely great, the series 
ws called an infinite series. 


Thus, if a series be formed by making any term the 
half of the preceding term and this process be continued 
indefinitely, as, 





a ws 1 1 EE a eee 
see eS S716) Ba. "2 
we have an infinite series. 


(6) If the sum of n terms of an infinite series can be 
shown to approach some finite number as n 1s made to 
approach infinity, the. series is called convergent. If this 
sum can not be shown to approach some finite quantity, the 
series is called divergent. 


In the discussion of the subject of series, we shall 
examine only three special forms, the arzthmetical series 
the geometrical series, and the binomial series. 


280 THE ESSENTIALS OF ALGEBRA 


Y. ARITHMETICAL SERIES (ArirumMeticaLt ProGREssion). 


201. Definition. An arithmetical series 1s a series in 
which the difference of any two successive terms 28 a constant. 


Illustrations. (1) 5, 9, 18, 17, 21, ---, is an arithmetical 
series, since the difference of any two successive terms is 4. 

(2) 8, 8.5, 4, 4.5, 5, 5.5, +--+, is arithmetical, since the 
difference is .5. 

(3) a,at+d,a+2d,a+5d,a+4d, --,a+(m—1)d, is 
arithmetical, since the difference of any two succes- 
sive terms is d. In this illustration @ is taken as any 
algebraic number, commensurable or not, and d likewise 
as any algebraic number. 


202. Notation. We shall denote by a the first term of 
any arithmetical series, by d the constant difference (com. 
mon difference) between any two successive terms, by / 
the last term or nth term, and by S' the sum of n terms of 
the series. 


203. Fundamental Formulas. 
(1) l=a+(m—1)d. 


(2) Sat xn 


= FAt aE xn, 


In these two relations five letters are involved, any two 
of which may be unknown. 

The first of the above formulas is easily seen to be true 
from the manner of formation of the general arithmetical 
series shown in illustration (8) above. 


SERIES. 281 


To derive formula (2), we write the series 
Ssat(atd)+(a+2d)+- panko! 
Then reverse the series, 
S=1+d—d)+(U—-—2d)+--+(a+2d)+(a+d)+a, 
and add the two equalities, giving 
28=(a+)+(a+)D+@4+)D+(a4+)+ 
+(a+l)+(a+l) 
= n(a+1), since there are n terms in the series. 


i genGek: l 
2 
The second form of formula (2) is derived by replacing 
1 by its value from formula (1). 


xn. 





204. Arithmetical Mean. If a, b,c be three successive 
terms forming an arithmetical series, 6 is called the arith- 
metical mean of a and e. 

=4(a+e), for by the definition of the arithmetical 
series, 


b—a=c—); 
(transposing), 2b6=a+e, 
or b=4(a+e). 


205. Arithmetical Means. In an arithmetical series a, 4, 
ce, d, e, f,--:,l, the terms J, ¢, d, e, f, +++, are called the arith- 
metical means ota and J. 


206. To insert k arithmetical means between any two 
numbers. 


Let a and 6 be any two numbers. After * means have 
been inserted, the whole series will consist of # + 2 terms. 


282 THE ESSENTIALS OF ALGEBRA. 


Hence, 6 is the last or (k + 2)” term of an arithmetical 
series of which a is the first, and d an unknown common 
difference. 





Hence, b=a+(k+2-l1)d 
=a+(k+1)d, 
Bs eed 
Ray 


The common difference d being known, the series may 
be easily written thus: 


9) Be 
a 44 20 a) 


bis eG; 
k+l eee 


a, 





207. An arithmetical series is determined when two of its 
terms are known. 


Let a be the kth and 6 be the mth term of an arith- 
metical series. Let a be the first term and y the common 
difference. 

Then a=x+(k—-1)y, 

b=x+(m—1)y. 
By subtraction 
b—a=(m—k)y, 
b— 


or v= oe (the common difference), 
m — 





by Ge a\| _(m—l)a—(k-1)@ 
and x=a—(k—1) {= iy [ = 
The first term 2 and the common difference y being 
known in terms of a, 6, &, m, the series may be written 
down. 
Determine the arithmetical series in which the 5th term 
ig 17, and the 12th term is 88. 


SERIES. 283 


SOLUTION. 
Let x = first term, and let y = common difference. 


Then =a-+ (n—1)d becomes respectively, 
17=x+(5-1)y, 
aus (12—-1)y. 
By subtraction, Bl =i y, OY ¥ = 3; 
when Ep ria 
Then the series is 5, 8, 11, 14, 17, 20, ---, 35, 38. 


EXERCISES. 
Find the 18th term of 2, 5, 7, 10, ete. 
Sum 4, 7, 10, etc., to 9 terms. 
Insert 5 arithmetical means between 10 and 34. 


PO B 


Find the 15th term of an arithmetical series whose 2d 
and 8th terms are 9 and 21, respectively. 


5. Which term of the series 1, 6, 11, 16 is 96? 

6. Find the sum of the natural numbers from 91 to 187. 

7. Show that if any four numbers are in arithmetical pro- 
gression, the sum of the Ist and 4th is the same as the sum of 
the 2d and 3d. 

8s. Find the 18th term of 27, 21, 15, 9, ete. 

9. Find the sum of 12 terms of 3, 44, 6, 74, ete. 

10. How many terms of 1, 2, 3, 4, etc., will make 465 ? 

11. How many terms of 7, 11, 15, 19, etc., will make 297 ? 

12. How many strokes does a clock strike in 12 hours ? 

13. Find the sum of all the even numbers from 100 to 200 
inclusive. 

14. Find the sum of all the numbers from 48 to 135 inclu- 
sive which are divisible by 3. 

15. What debt could be paid in a year by the payment of 
10 ¢ the 1st week, 40¢ the 2d week, 70 ¢ the 3d week, etc.? 


284 THE ESSENTIALS OF ALGEBRA. 


16. Determine the series whose 10th term is 51 and whose 
20th term is 101. 

17. Determine a series whose 15th term is 0 and whose 31st 
term is 64. 

18. Find the sum of all numbers from 105 to 361 inclusive, 
which, when divided by 4, leave a remainder of 1. 


Il. GEOMETRICAL SERIES (GEOMETRICAL PROGRESSION). 


208. Definitions. A geometrical series is a series in which 
the ratio of any two successive terms ts a constant. 

Illustrations. (1) 2, 4, 8, 16, 32, is a geometrical series 
in which the ratio 16+8=8+4=32+16=2 is a constant. 

(2) J, 4, 4, a gh +) 18 a geometrical series with ratio 
equal to 4. 

(3) 1, x, 2*, 23, at, ---, is a geometrical series with ratio 
equal to z. 

(4) a, ar, ar’, ar, art, ---, ar""|, is a geometrical series 
in which the ratio is 7. 


209. Notation. ‘The illustration (4) above suggests a 
notation for the geometrical series. 
(1) a = first term. 
(2) r =constant ratio. 
(3) J = last term, or nth term. 
(4) S=sum of n terms. 
210. Formulas. 
CO eat oe 


(2) S=a("=7)=a(5=") 


(3) S= j “_, when r <1, and n=a, 
—r 











SERIES. 285 


The first of these formulas results from the law of forma- 
tion of the series as indicated in illustration (4) above. 
The second formula we may derive as follows: 


GQ) S=a+ar+ar4+ ar+---+ ar? + art 
Multiply by 7, 

(2) Sr=ar + ar? + are +--+ +ar"* + ar”! + ar". 
Subtracting (2) from (1), 

(3) S—Sr=a-— ar". 

(4) SQ-r)=ad—>2). 


(5) ieee x cz “). 





—?r 
Another method of derivation is worthy of attention. 
By actual division we know that 











ats: 
1 pee te 71 72, 

l-—r 

Seid 

oe (a pte 
1l—r 

eu, 5 

I Peltr4r4r4+7, 
1l—r 


and so on; for the general case, 


Foal rr rt oe bm 
4 8 eet i 


Now by writing the value of S again, 
S=a+ar + ar? + ar? + art + ee + ar™? + ar, 
and factoring out a from each term on the right, we have 
S=a(l trp ret ro foe for? rr), 
The value in this bracket is the same as the value of 


above; hence, Pa 
Sean. ( ) 


et 2” 





ee et 





me fe 





286 THE ESSENTIALS OF ALGEBRA. 


211. Sum to Infinity when r<1. If the ratio 7 be less 


: a 
than unity, r°<1,and whenn=o,7r"=0. Hence;s= f 

| ; —r 
when r< ll, andn=o 





The sign = is read approaches. 
Illustration. Find the sum of 1+4+4+ $4 ete. 


1 
i ae 





1 
Here, a=1,r= = and S'= 
_ 


212. The Geometrical Mean. Jf a, 6, c, be three successive 
terms of a geometrical series, then 6 is equal to the square 
root of the product of a by c, and is the geometrical mean of 
a and e. 


By the definition of a-geometrical series, 
b 


c — 
Fare. whence, 6? = ac, or b = Vae. 
a 


213. Geometrical Means. Jn a geometrical series the 
terms lying between any two terms are called the geometrical 
means of those two terms. } 

Thus, 5, 10, 20, 40, 80, 160, are six terms of a geometrical 
series with ratio 2. The terms 10, 20, 40, 80, are the four 
geometrical means between 5 and 160. 


214. Insertion of Geometrical Means. Any number of geo- 
metrical means may be inserted between any two numbers. 


Proof. Let aand 6b be any two numbers, and let k be 
the number of means to be inserted. Then 6 is the 
(k+2)” term of a geometrical series, whose first term 
isa. Ifr be the unknown ratio, 

Writ ipl hinge cmengs Or 


ip eee slog 
sae" 


SERIES. ei ORF 


Insert five geometrical means between 128 and 2. 
These means may be written down if we know the value 
of the ratio. ‘This is given by 


pat N{5, where = 5, b= 128, a=2, 
a 
" AL Vv 64 = 2. 


Hence, the series is 2, 4, 8, 16, 82, 64, 128, and the 
means are 4, 8, 16, 32, 64. 


215. A Geometrical Series Known. <A geometrical series 
is known when any two terms are known. 

Proof. Leta be the &th, and 7 the mth terms of a geo- 
metrical series; and let 7 be the unknown ratio, with zx 
as first term. 

Then a= zr'-}; 
and Pere. 

These two equations are sufficient to determine the first 
term x and the ratio r. 


‘ ps ey 1 
By division —— =-, 
Co ee 
or prern. &, 
a 
m—k l 
(1) ° 7 = i, 





Equations (2) and (1) give the’ first term and ratis, 
respectively, of the required series. 


288 THE ESSENTIALS OF ALGEBRA. 


EXERCISES. 


1. Find the 10th term of 1, 2, 4, 8, etc. 
2. Find the sum of the 10 terms in (1) above. 


3. The 5d and 6th terins of a geometrical series are 27 and 
729. Find the 8th term and the sum of the 8 terms. 


4, Find the sum of 10 terms of 1, 4, 4, 4, ete. 
5. Sum 3, — 3’, 3°, — 3‘ to 8 terms. 


6. A house with 8 windows was sold for $1 for the 1st 
window, $2 for the 2d, $4 for the 5d, ete. What was received 
for the house ? 

7. If you receive $5 Jan. 1, $10 Feb. 1, $20 March 1, and 
so on for each month of the year, what is the total amount you 
will receive during the year ? 





8. Find the sum to infinity of 535, 735, zara ete. 
9. Find the sum to infinity of 1, 4, 4, 34, ete. 


10. The arithmetical mean of two numbers is 13 and their 
geometrical mean is 12. Find the numbers. 


11. Show that the series of alternate terms of a geometrical 
series is also a geometrical series. 


12. If every term of a geometrical series is divided by the 
same quantity, the quotients form a geometrical series. 


13. The reciprocals of the terms of a geometrical series form 
a geometrical series. 


14. The difference between the 1st and 4th of four numbers 
in geometrical progression is 208, and between the 2d and 3d 
is 48. Find the numbers. 


15. The sum of 3 numbers in geometrical progression is 14 
and the sum of their reciprocals is 7. Find the numbers. 


a 


SERIES. 2Q8y 


Ill. BINOMIAL SERIES. 
216. Definition. The series defined by 


(a + b)® =a" tna + ey LD gr-2h2 pose 


4 a ae 3) oe (M=P ED) nrg. 
T 
is called the benomial series. 
Nore. In higher algebra it is shown that this series gives a true 


value of (a+6)” for all values of a and b provided n is an integer. It 
also defines (a +b)" properly when n is a negative number, or a 


fraction, provided ® bea proper fraction. The general proof of these 
a 


assertions will not be attempted in this development. 


EXERCISES. 


ieee inthis n= 30, ¢— 1, b'= 2. 





(1+ 2)=143004 0%) wg AXES a® + ete. 


2. Write out the first 6 terms of (1+2)"; (1—y)"; (w+y)” 
3. Find the coefficient of 2” in (1+). (Referring to the 
binomial series, we see that 7 = 15.) 
20 0 CS Co ara a Ga 
— 3)(—3—1)(—3— 2) 
a rca v ea ni (— 2) 





=14+-32+4+627 + 102°- ete. 
5. Write out the first 5 terms of 


(a) A—2)7; (&) +2); © G—-a)?; @ A+ay. 


290 THE ESSENTIALS OF ALGEBRA. 


EXERCISES— MISCELLANEOUS. 
1. Expand (w#+2y)°; (2a—y)*®; (2a—38b)). 
2. Find the sum of 8 terms of 1, 2a, 42%, «+. 
3. Find the sum of 30 terms of 7, 11, 15, «+». 
4. Find the 6th term of (1—2a)~*. 


5. How many arithmetical means must be inserted between 
10 and 40 so that the sum of the series may be 275? 


6. Divide 26 into three parts which are in geometrical pro- 
gression, and such that when 4 is added to the second part, the 
three parts are in arithmetical progression. 


7. Insert a geometrical mean between 5 and 45. Explain 
the double sign of the result. 


8. Sum to infinity 3, 2, 38, +. 
9. Find the coefficient of ay’ in (24—3 y)™ 
10. Find the coefficient cf a in (1 — 2a)-?, 


11. In how many ways can 9 persons be selected from a 
party of 21 people? 


12. How many commiitees consisting of 4 men and 3 women 
can be formed from 12 men and 10 women? 


13. Each member of a baseball nine, except pitcher and 
catcher, can play in any position. In how many ways can the 
team be arranged upon the field ? 


14. In how many ways may a baseball nine be selected from 
16 candidates, if 2 are pitchers, 3 are catchers, and the remain- 
der can play in any position ? 


15. Expand (4 — 23 to five terms, and thus get an approxi- 
mate value of 2. 


SERIES. 291 


16. Expand (100-1)? to five terms, and thus get an approxi- 
mate value of 99. 


a 6 
17. Expand (1 +0) . 


1s. Find the sum of all the even numbers between 205 and 
341. 
19. Find the sum to infinity of 1 ++ +5 t + —— 1 — -[- ove, 
V2 2/2 
_ 20. The sum of the first three terms of a geometrical series 
is 21, and the sum of their squares is 273. Find the series. 


21. A debt is to be paid by 10 payments which form an 
arithmetical progression. The third payment is $220, and the 
seventh is $360. Find the last payment and the total debt. 


22. How many consecutive odd numbers beginning with 11 
must be taken to make a sum of 759? 


23. Prove that the squares of the terms of a geometrical 
series also form a geometrical series. 


24. (Va+1—Va—1)'= what? 
25. Expand (1+2+4 27). 


CHAPTER XX. 
LOGARITHMS. 


217. Definition. The exponent with which a given num- 
ber a must be affected in order to produce another number N 
ts called the logarithm of N to the base a. 


Hxamples : 

(1) The logarithm of 32 to the base 2 is 5, for 2°=382. 
(2) The logarithm of 125 to the base 5 is 3, for 5°=125. 
(3) The logarithm of 100 to the base 10 is 2, for 107= 100. 


218. Notation. log 


oa 


N means the logarithm of WV to the 
base a. 
The two equations, 


a*= N and x=log, NV are’ equivalent. 


219. System of Logarithms. The logarithms of all 
numbers to any given base constitute a system of loga- 
rithms. Any positive number other than unity may be 
used as the base. Hence, there may be an infinite number 
of logarithmic systems. ‘There are only two systems in 
common use, the Napiertan and the Common. ‘The Napier- 
ian system uses the base e (2.7182818). This system is 
used in analytical and theoretical discussions and is but 
rarely used in caleulation. The Common or Briggs system 
uses the base 10. ‘This system is the one in general use. 

292 


LOGARITHMS. 293 


220. Properties of Logarithms. 
(1) The logarithm of 1 is 0. 


Since a°=1 for all finite values of a, log 1=0 for all 
bases. 


(2) The logarithm of the base itself is unity. 


Since tea, lov a = 1. 


(3) The logarithm of a product is equal to the sum of the 
logarithms of its factors. 

Let F, and F, be the factors, and put #, = a" and F, = a’. 
Pieter, and y—log, i. Fx £,=ata =a. 
log, (Ff, x £,) =2+ y= log, Ff, + log, J. 

Similarly, log, (f’, x F, x #3) =log, #, + log, F,+log, F;. 

The property is true for any number of factors. 

Ex. log,, 210 = log,, (2x 3 x0 x 7) =log,, 2 + log,, 3 
+ logy 5 + log, 7. 


Ss 


(4) The logarithm of a quotient is equal to the logarithm 
of the dividend minus the logarithm of the divisor. 

Let #, and F, be dividend and divisor respectively and 
represent them as in (3). 


> 


log, (3) =z—y=log, lf, —log, F,. 


BD semen eS hast ~) 
Ex. logy 13 a log 49 35 logy 13: 


ah pee eT. ou tes 
_ logy, 5 + logy 7 logy 2 logy 2—log,, 3. 


294 THE ESSENTIALS OF ALGEBRA. 


(5) The logarithm of the nth power of a number is equal 
to n times the logarithm of the number. n may be integral 
or fractional. 

If WV” be the given power, 
put IViexnges 
then, Nee 

log, NV" = nz = nlog, WV, since — oma 
Ex: log,,310% =o lop nolo. 
logy) V125 = logy 125% = 4 logy 125. 


221. Characteristic and Mantissa. 
The equation for Common Logarithms is 107 = WN. 
This equation shows that the value of 2 (the common 
logarithm) will not in general be integral and that it may 
be positive or negative. 
$756 > 10°sand <a10 


Hence, log 8756 = 3 + a fraction. 
058 > 10-2 and 10 3 
Hence, log .053 = — 2 +4 a fraction. 


The Characteristic is the integral part of a logarithm. 
The Mantissa is the fractional part of a logarithm. 


222. To find the Characteristic : 
(1) Of the logarithm of a number greater than unity. 


LOO ein 
0} ech) 
1.07100 
103 = 1000 


10" = 1 with n ciphers annexed. 


LOGARITHMS. 295 


A study of the above table shows that the logarithm of 
a number between 1 and 10 is 0+a fraction; between 
10 and 100, 1+ a fraction; between 100 and 1000, 2+a 
fraction; and so on. 

If V is a number whose integral part has n digits, we 


“may write N a 190@-D+ a fraction 
Hence, log V = (n—1)+ a fraction. 


The characteristic of the logarithm of a number greater 
than unity ts one less than the number of digits in its 
entegral part. 


Ex. log 8542 = 3 + a fraction. 
log 96.54 = 1 + a fraction. 


(2) Of the logarithm of a decimal fraction. 


Aer ete 
tulle Or 4 
OU tree UEs 
S000 Peers 1. OQ 


.(n ciphers) 1 = 10-@+» 


An inspection of the above table shows that the logarithm 
of anumber between 1 and .1 is —1+ a fraction; between 
-Land .01 is —2+a fraction; between .01 and .001 is —3+ 
-afraction; between .001 and .0001 is —4+a fraction; of any 
decimal beginning with n ciphers is —(n+1)+a fraction. 

Let V = a decimal fraction beginning with n ciphers. 


Then, N = 10-@+) +2 fraction, 
log MV =— (n+ 1) + a fraction. 


296 THE ESSENTIALS OF ALGEBRA. 


The characteristic of the logarithm of a decimal fraction is 
one more than the number of ciphers immediately after the 
decimal point, and is negative. 


Ex. log .0856 = — 2 + a fraction. 
log .00058 = — 4 + a fraction. 


EXERCISES, 


Determine the characteristic of the logarithm of each 
of the folowing numbers: 


Dips rs ire AL MAS Lovo 7. .0000031. 
2. 95.64. Breas 8. 231.416. 
3, aD: 625.005 (0 9. (63454. 


10. 1375.603. 


223. The mantissa depends upon the sequence of digits, 
and not upon the position of the decimal point. 
The logarithm of 4880 is 3.6415. 
10345 = 4380 or log 4380 = 3.6415. 


Dividing in succession by 10, 
1025 — 488. or. Jop 436 =e 
1015 — 43:8 or log 43.8 == JeGci. 
10°45 = 4.38 or log 4.38= 0.6415. 
10-145 = 4388 or log .438=—1+4.6415. 
10-2+-6415 — 0438 or log .0488 =— 2+ .6415. 
- 10-8+ 6415 — .00438 or log .00438 = — 3 + .6415. 


This table makes it evident that the mantissa does not 
change so long as the sequence of digits is unchanged. 


LOGARITHMS. 297 


EXERCISES. 
log 456 = 2.6590; what is log 456000 ? 
log .987 = —1+4.9943; what is log 9870? 
log 57.9 = 1.7627; what is log .000579? 
log 3210 = 3.5065; what is log 3.21? 
log .0705 =— 2+ 8482; what is log 7050000? 
log 7.350 is .8663; what is log 73500 ? 
log 2 is .8010; find the log 2%, 24, 2°. 
log 3 is .4771; find log 3%, 3%. 
From examples 7 and 8 find the log 6; log 24. 
log 5 is .6990; find log 25, log 125. 
From data above given find log 20, log 75. 
. log Tis .8451; find log 49, mg 343. 
Find log 42. 


Find log 140. 
reed a x oF 


Oe Oe Se eee 


KH YH eH eS eH 
Piet OTe RE tN 


mn 


Find log 


USE OF TABLES. 


224. Approximate Arithmetical Calculations involving 
multiplications, divisions, raising to powers, and extraction 
of roots may be greatly abridged by the use of Logarith- 
mic Tables. Such tables have been constructed showing 
the Common Logarithms of numbers from 1 to 10,000. 
We give a four-place table showing the mantissas of the 
logarithms of numbers from 1 to 1000. 


THE ESSENTIALS 


OF ALGEBRA. 








O 


1 


Det 





3 





4 





5 6 7 8 9 















































































































0000 0043 0086 0128 0170 | 0212 0253 0294 0834 0874 
11 | 0414 0453 0492 0531 0569 | 0607 0645 0682 0719 0755 
2 | O792 0828 0864 0899 0934 | 0969 1004 1038 1072 1106 
13 | 1189 1173 1206 1239 1271 | 1303. 13337 “Jae; eae 
14 | 1461 1492 15293 1553 1584 | 1614 1644 1673 "u70sea7ee 
15| 1761 1790 1818 1847 1875 | 1903 1931 1959 1987 2014 
16 | 2041 2068. 2005 2122 2148 | 2175 22901 2997) 9953) 99070 
17 | 2304 2330 ~ 2355 2380 2405 | 2480 2455 2480 2504 9599 
18 | 2553 2577 2601 2625 2648 | 2672 2695 2718 2742 2765 
19 | 2788 2810 2833 2856 2878 ; 2900 29238 2945 929067  ~90R9 
20; 3010 3032 3054 38075 3096 | 3118 3139 3160 3181 3201 
Y1 | 3222 3243 3263 3284 3304 | 3324 3845 3365 3385 3404 
29 | 3494 3444 3464 3483 4502 | 3592 3541 3560 3579 3598 
23 | 3617 3636 3655 3674 3692 | 3711 3729 3747 VaiGeuneres 
24 | 3802 3820 3838 3856 5874 | 3892 3909 3927 3945 3962 
25 | 3979 38997 4014 4031 4048 | 4065 4082 4099 4116 4133 
26 | 4150 4166 4183 4200 4216 | 4232 4249 4265 4281 4298 
97 | 4314 4330 4346 43862 4878 | 4893 4409 4425 4440 4456 
98 | 4472 4487 4502 4518 453: 4548 4564 4579 4594 4609 
99 | 4624 4639 4654 4669 4683 | 4698 4713 4728 4742 4757 
30 | 4771 4786 4800 4814 4829 | 4843 4857 4871 4886 4900 
3 4914 4928 4942. 4955 4969 | 4983 4997 5011 5024 5038 
32 | 5051 5065 5079 5092 5105 | 5119 5132 5145 5159 5172 
33. |°5185 5198 5211 5224 52937 | 5250 5263 5276 5289 5302 
34 | 5315 5328 5340 5353 5366 | 5378 5391 5403 6416 5428 
35 | 5441 5453 5465 5478 5490 | 5502 5514 5527 5539 5551 
36 | 5563 5575 5587 5599 5611 | 5623 5635 5647 5658 5670 
37 | 5682 5694 5705 5717 5729 | 5740 5752 5763 5775 5786 
38 | 5798 5809 5821 5832 5843. | .5855 5866 5877 5888 5899 
39 | 5911 5922 5983 5944 5955 | 5966 5977 5988 5999 6010: 
40 | 6021 6031 6042 6053 6064 | 6075 6085 6096 6107 6117 
41 | 6128 6138 6149 6160 6170 | 6180 6191 6201 6212 6222 
2 | 6232 6243 6253 6263 6274 | 6284 6294 6304 6314 6825 
3 | 6835 6345 6355 6365 6375 | 6385 6395 6405 6415 6425 
44 | 64385 6444 6454 6464 6474 | 6484 6493 6503 6513 6522 
45 | 6532 6542 6551 6561 6571 | 6580 6590 6599 6609 6618 
46 | 6628 6637 6646 6656 6665 | 6675 6684 6693 6702 6712 
47 | 6721 673 6739 6749 6758 6767 6776 6785 6794 6803 
48 | 6812 6821 6830 6839 6848 | 6857 6866 6875 6884 6893 
49 | 6902 6911 6920 6928 6937 | 6946 6955 6964 6972 6981 
50 | 6990 6998 7007 7016 7024 | 7033 7042 7050 7059 ‘7067 
51 | 7076 7084 7098 7101 7110 | 7118 7196 “V13h Jide 
yn Aaa MrGiaeh. WON” <item 7202 7210 7218 7226. ssieee 
53. | 7243 7251 . 7259 7267 7275 | 7284 7292 7300 . 7308 ue 
54 | 7324 7332 7340 7348 7356 | 7364 7372 7380.) 7oSeeue 
N O 1 2 3 4 5 6 v4 8 9 





















































LOGARITHMS. 299 

N]/] O 1 2 3 4 5 6 7 8 9 

55 | 7404 7412 7419 7427 7435 (443 7451 7459 7466. 7474 
56 7482 7490 7497 7505- 7513 TOLD, Des) TbS6r' (ozo UL DDL 
oT 7559 7566 7574 7582 7589 7597 7604 7612 7619 7627 
58 7634 7642 7649 7657 T664 7672 T679 7686 7694 £7701 
59 T1709 7716 7723) T7310 1738 | T1745 «T1752. «7760)=— TT6T)~—s T7174 
60 | 7782 7789 7796 7803 7810 (818 7825. 7832) 7839 7846 
61 7853 7860 7868 7875 7882 7889 7896 7903 7910 7917 
62 7924 7931 ° 7988 7945 7952 7959 7966 7973 7980 T7987 
63 7993 8000 8007 8014 8021 8028 80385 8041 8048 8055 
64 8062 8069 8075 8082 8089 8096 8102 8109 8116 8122 
65] 8129 8136 8142 8149 8156 8162 8169 8176 8182 8189 
66 8195 8202 8209 8215 3292, 228), (S200 |) S241 8248 8254 
67 8261 8267 8274 8280 8287 8293 8299 8306 8312 8319 
68 8395 8331 83838 8344 8351 8357 83863 8370 8376 £8382 
69 8388 8395 8401 8407 8414 8420 8426 8432 8439 8445 
7O)} 8451 8457 8463 8470 8476 8482 8488 8494 8500 8506 
V1 8513 80919 8525 8531 853 8543 8549 8555 8561 £8567 
72 8573 8579 8585 # £8591 597 8603 8609 8615 £8621 8627 
73 8633 $8639 8645 8651 8657 8663 8669 8675 8681 8686 
74 8692 8698 8704 8710 8716 SLO hk IEOloon (OlOo. 6 O40 
75 )| 8751 8756 8762 8768 8774 8779 8785 8791 8797 8802 
76 8808 8814 8820 8825 8531 8837 8842 8848 8854 8859 
Lit 8865 8871 8876 8882 8887 88938 8899 8904 8910 8915 
78 8921 8927 89382 8938 8943 8949 $8954 8960 8965 8971 
79 8976 8982 8987 8993 8998 9004 9009 9015 9020 9025 
80 | 9031 90386 9042 9047 9053 9058 9063 9069 9074 9079 
sl 9085 9090 9096 9101 9106 9112 117 9122 9128 9135 
82 O1388 °° 9143 9149 9154 9159 9165 9170 9175 9180 9186 
83 9191 9196 9201 9206 9212 OO Ty: 92225 9227) 19232. 9235 
84 0243 9248 9253 9258 9263 9269 9274 9279 9284 9289 
85 | 9294 9299 9304 9309 9315 9320 93825 9330 93835 9340 
86 | 9345 9350 9355 9360 9365 | 9370 9375 9380 9385 9390 
87 | 9395 9400 9405 9410 9415 | 9420 9425 9430 9435 9440 
88 | 9445 9450 9455 9460 9465 | 9469 9474 9479 9484 9489 
g9 | 9494 9499 9504 9509 9513 | 9518 9523 9528 9533 9538 
90 | 9542 9547 9552 9557 9562 9566 9571 9576 9581 9586 
Ol 9590 9595 9600 9605 9609 9614 9619 9624 9628 9633 
92 9638 9643 9647 9652 9657 9661 9666 9671 9675 9680 
93 9685 9689 9694 9699 9703 9708 9713 9717 9722 9727 
94. 9731 97386 9741 9745 £9750 9754 9759 9763 9768 9773 
951 9777 9782 9786 9791 9795 9800 9805 9809 9814 9818 
96 9823 9827 9832 9836 9841 9845 9850 9854 9859 9863 
O7 9868 9872 9877 9881 9886 9890 9894 9899 9903 9908 
98 9912. 9917 9921 9926 9930 9934 9939 9943 9948 9952 
99 9956 9961 9965 9969 9974 9978 9983 9987 9991 9996 
N O 1 2 3 4 5 6 7 8 9 


300 THE ESSENTIALS OF ALGEBRA. 


225. To find the Logarithm of a Given Number. The 
logarithm of any number consisting of three consecutive 
digits has its mantissa given in the table, and its charac- 
teristic is found by Art. 222. 

Hxamples: 

(1) Find log 358. } 

The first two digits, 35, are found in the column marked 
NV, and the third digit, 8, is found in the top row. 

In the row opposite 35, and in the column under 8, we 
find the number 5539, which is the mantissa of the logarithm 
of 858. By Art. 222, the characteristic is 2. 

Hence, log 308 = 2.5539, 


(2) Find log 755. 
In the row opposite 75, and in the column under 3, we 
find 8768. The characteristic is 2. 


Hence, log 753 = 2.8768. 


(3) Find log 7.53. 
The mantissa is the same as for 753, and the character- 


Hence, log 7.55 = 0.8768. 

(4) Find log .0758. 

The mantissa is the same as in (2) and (8), and the 
characteristic is — 2. 

Hence, log .0753 = — 2+ .8768. 


This is usually written 2.8768. 
The negative characteristic is frequently made positive 
by adding and subtracting 10. 


Thus, log .0753 = 2.8768 = 8.8768 — 10. 


LOGARITHMS. 301 


The characteristic may be retained as a negative, but 
the mantissa is usually written positive, and is not sub- 
tracted from the negative characteristic. 


(5) Find log 3965. 

The mantissa for 396 is .5977 and for 397 it is .5988. 
The difference is .0011. The mantissa for 3965 is approxi- 
mately .5977 + .5 x .0011 = .59825. Since we are to re- 
tain but four places, this would be written .5983. The 
characteristic is 3. 

Henee, log 8965 = 3.5983. 


By means of such approximations as above shown, the 
accompanying table may be used to find quite accurately 
the logarithms of numbers to 10,000. 


(6) Find log 87.55. 

The mantissa for 875 is .9420, and for 876 it is .9425. 
The difference is .0005. ‘The mantissa for 8755 is approxi- 
mately .9420 + .3 x .0006=.9422. The characteristic is 1. 

Hence, log 87.53 = 1.9422. 


226. To find the Number corresponding to a Given Loga- 
rithm (Antilogarithm ). 

Hxamples : 

(1) Find the number whose logarithm is 2.6877. Find 
in the table the given mantissa or the one nearest to it, 
but less. | 

5877 is found in the row opposite 38 and in the column 
under 7. Hence, the three digits of the number are 587. 
Since the characteristic is 2, the number is 387. 

The number whose logarithm is 2.5877 is 387. This is 
frequently written log~! 2.5877 = 387. 


302 THE ESSENTIALS OF ALGEBRA. 


(2) Find the number whose logarithm is 1.8963. 

In the table the nearest mantissa to 8963 is 8960. 8960 
corresponds to the sequence of digits 787. ‘The difference 
between 8960 and the given mantissa is 8. The difference 
between 8960 and the following mantissa in the table is 5. 
3+5=.6. The next digit in the sequence is 6. The 
sequence of digits corresponding to the mantissa 8963 is 
7876. ‘The characteristic 1 shows that there must be two 
digits in the integral part. 


Therefore, log-) 1.8963/==(3eia. 


(3) Find the number whose logarithm is 3.5896. 

The mantissa in the table next below 5896 is 5888, 
which corresponds to the sequence of digits 388. The 
difference between 5888 and the given mantissa is 8. The 
difference between 5888 and the following mantissa in 
the table is 11. 8+11=.7. This gives 7 as the fourth 
digit of the sequence. Since the characteristic is — 3, 
the first significant figure of the required number is in 
the third decimal place. 


Therefore, log! 3.5896 = .003887. 


EXERCISES. 


Find from the table the logarithms of each of the 
following numbers: 


Load. 5, 320.3, 9. .00056. 
2. 984. 6. 948.6. 10. .5236. 
3. 506. Be lkaavilie 11. 3.1416. 
4. 1345. 8. .0679. 12) 2 


LOGARITHMS. 303 


13, 1.40. 16. 3. 19. .3964.. 
14. 122. 17. 984}. 20. .005876. 
15. 37}. ; 18.79.0906. 


Find the number corresponding to each of the following 
logarithms : 


21. 2.8136. 28. 3.2084. 35. 9.4669 — 10. 
22, 1.9145. 29. 0.3756. 36, 1.4784. 

23. 2.8904. 30. 1.5967. ST leon, 
eae, (001, Bie pio, ag. 97/38945'— 10. 
rime tna yee g28eo1 584: 39. 9.4896 — 10. 
26. 0.4472. 3405, 5922. 40. 8.5986 —10. 
OTe ako iO. 34. 8.6493 — 10. 


227. To perform Arithmetical Operations by the Use of 
Logarithms. 


Examples : 
(1) Find the value of 469 x 541 x .0878. 
By Art. 220 (3), the logarithm of a product is equal to 
the sum of the logarithms of the factors. 
log 469= 2.6712 
log 541= 2.7332 
lope 0818 =~ 8.5775 — 10 
log of product = 13.9819 — 10 = 3.9819 


The product = log? 3.9819 = 9592. 


304 THE ESSENTIALS OF ALGEBRA. 


5237 x Gi9dee 0059 


By orinditherealie at 
C2) HANG he alte Ob Gas ne 





By Art. 220 (4), the logarithm of a quotient is equal to 
the logarithm of the dividend minus the logarithm of the 


divisor. 
169025 (et oe 


log 6.94= 0.8414 
log .0059= 7.7709—10 


Adding, log of numer- = 12.3314—10=12.3314—-10 (1) 
ator (dividend) 


log 497 = 2.6964 
log .7648 = 9.8836 — 10 
lop oF = 0S1G58 
Adding, log of de- | 
aay = 13.7159 —10 =i 9 
nominator (divisor ) j : (4) 


Subtracting (2) from (1), log of quotient = 9.5555 — 10 





The quotient = log~!(9.5555 — 10) = 0.85983. 


V578 x .038575 x V38.7 
292 x V.875 x V8.58 








(3) Find the value of 


We apply the principles of Art. 220. 


log V578 = 4 log 578=4 x2.7619= 1.3809 
log .03575= 8.5533—10 
log V38.7 =1 log 88.7=1 x 1.5877= 0.5292 
Adding, log of numerator = 10.4634—10 (1) 





LOGARITHMS. 308 


log 29? = 2 log 29 = 2 x 1.4624=2.9248 
log V.375 = 4 log .875 


= 4x (9.5740 —10) =4.7870—5 


log V3.58 = d1og 3.58 =1 x 0.5539 = 0.1846 





Adding, log of denominator =7.8964—5 
= 2.8964 (2) 
Subtracting from log of dividend 10.4634 —10 (1) 
log of divisor 2.8964 (2) 





log of quotient = 7.5670 — 10 


The quotient = log !¢7.5670—10)= .00569. 


EXERCISES, 


By the use of logarithms evaluate the expressions given 


below : 


a 


nt ee 


6. 


487 x 394 x 3.75. 

57.84 x 589 x 427.5 x 52.36. 
oTd4 x 69.54 x 16.66 x .000584. 
17.58 x 125.6 « .00825 x 2.65. 
1.578 x 3584 x 27.5 x .398. 


3857 x 47.65 x 394, 
467 x 89.84 x 3.875 





7, BA416 x 13.6% x 87.5 9 -V946 x 39.872 x V4.749 


57.6 x 39.57 
.0236 x 34.63 





V84.65 x 8.75% x 39.46 
10. V658 x 47.688 x V380.57. 





" 48.94 x 3.92. 4982 x .0388 x .00051 


228. Exponential Equations. Hquations in which the 
unknown quantity occurs as an exponent are called expo- 
nential equations. ‘They are easily solved by logarithms. 


306 THE ESSENTIALS OF ALGEBRA. 


Ex. Find the value of a in 267 = 137. 
Taking the logarithms of both members of the equation, 
ars log 25° = log 137. 
x log 25 = log 187. 


Plog oie od 





Se isa oa sara 1.5285. 
EXERCISES, 
Solve the following equations : 
Lael 6. Cleats": 
Cue BX e100, 7... (00 aes 
oy daub ap 8. (OC) * aa 
Seo te Ope 9. 108 e*— ieee 
Se ee 10. 3" 'X 574 


229. Compound Interest. The computation of com- 
pound interest may be easily performed by the use of 
logarithms. : 


Let P=principal loaned. 
# = rate of interest. 
n = time in years. 
A,, = compound amount for n years. 
Then A,=P+ PR=P(1+ £R), amount for one year. 
A,=PA1+Rh)+P14+ 8)R=PA+ &)?, 


amount for two years. 


A, = P(1+ R&)", amount for n years. 


LOGARITHMS. 307 

If the interest be compounded semi-annually, the com- 
pound amount is R\2 
a=r(te¥ 


If compounded quarterly, it is 


A, 2 PU i zy". 


Examples: 

(1) Find the amount of $850, compounded annually 
for 8 years, at 5%. : 
Ag= S00 (1+ .05)? = 350(1.05)° 

log 850 = 2.9294 
8 log (1.05) = 0.1696 
log A, = 3.0990 

A, = log! 3.0990 = $ 1256. 





(2) Find the amount of $598, compounded semi- 
annually for 6 years, at 4%. 


PVs bet @ Deg be, 
log 598 = 2.7767 
12 log 1.02 = 0.1032 
log A, = 2.8799 
A, = log™1 2.8799 = $758.40. 





(3) In how many years will $596, compounded annually 
at 4%, amount to $978 ? | 
From the general formula 
Ae Cl fh), 


we have GLE i, t= oe 


308 THE ESSENTIALS OF ALGEBRA. 


n log (1+ &)= log A, — log P 
" lop A; loge 
log(1+ &) 


Substituting the values of A, P, and R, we have 
_ log 978 — log 596 
log (1.04) 


2.9903 — 2.7752 
__ 2:9903 — 2.71702 _ 49 6 
0.0170 PA 


EXERCISES. 


1. Find the amount of $1565, compounded annually 
for 10 years, at 44%. 
2. Find the amount of $1968, ere seml- 


annually for 12 years, at 4%. 


3. In what time will $965 amount to $1500 when 
compounded annually at 33% ? : 


4. What principal, compounded annually for 20 years _ 
at 34%, will amount to $2500 ? 


5. Find the time in which a given sum will double 
itself when compounded annually at 3%. 


6. Find the compound amount of $1 at 4%’for one 
century. 


7. Find the rate at which a given sum will double itself 
in 20 years, if compounded annually. 


8. Find the amount of $364, compounded annually for 
12 yr. 5 mo. 10 da., at 5%. 


INDEX. 


Abscissas, axis of, 148, 149. 

Addition, 20-28, 35; identities, 26, 
27; of fractions, 117, 118; in 
elimination, 161. 

Algebraic expression, 4; signs used 
in, 5, 6; addition, 20, 21; sub- 
traction, 28, 29; fraction, 109. 

Arithmetic, numerical, 3; literal, 3, 
4; addition in, 20; square root 
in, 180, 181; cube root in, 184. 

Arithmetical mean, 281. 

Arithmetical means, 281. 

Arithmetical series (progression), 
280. 

Associative law, definition of, 23; 
of factors, 45. 

Axioms, 16, 


Binomial, definition of, 25 ; identi- 
ties, 70, 71, 74, 75; theorem, 76 ; 
factors, 100; series, 289. 


Coefficients, definition of, 7, 8; de- 
tached, 63; law of, 77; of a 
quadratic, 222. 

Combinations, examples of, 274; 
definition of, 275; symbol of, 
275. 

Commutative law, definition of, 22 ; 
of factors, 43. . 

Constants, 15, 125. 

Coordinate axes, 148, 149. 

Cube root, 182, 184. 


Distributive law, of factors, 45. 
Division, 58-67 ; identities, 66, 81; 
of fractions, 122. 


Elimination, definition of, 157; in 
substitution, 158; by compari- 
son, 160 ; by addition or subtrac- 
tion, 161; illustrative examples 
in, 162-164, 

Equation, definition of, 14; sign of 
15 ; root of, 16, 126 ; conditional, 
125; linear, 125, 147, 148, 170, 
229; of second or higher degree, 
140; fractional, 141, 148; inte- 
gral, 143; simultaneous, 157- 
173 ; quadratic, 207-228 ; homo- 
geneous, 239; symmetrical, 241 ; 
in higher degrees, 246. 

Evolution, 174-186. 

Exponents, definition of, 5; law of, 
77; integral, 187. 


Factoring, iii, 83-103. 

Factors, definition of, 42, 83; laws 
governing, 43-51; degree and 
number of, 48; monomial, 84; 
by rearrangement and grouping, 
97-99; binomial, 100; rational- 
izing, 198. 

Fractions, in number system, 2; 
algebraic, 109; signs of, 110, 
111 ; reduction of, 111, 115, 116; 
proper and improper, 114 ; addi- 


309 


310 


tion and subtraction of, 117, 118 ; 
multiplication of, 119, 120; divi- 
sion of, 122 ; complex, 128. 


Geometrical series, 284-287. 

Graph, the, iii; of the linear equa- 
tion, 148; of two linear equa- 
tions, 155; of surds, 192, 193 ; 
of the quadratic, 218-221. 


Highest common divisor, 104, 105. 
Highest common factor, 104. 


Identity, definition of, 14, 125; sign 
of, 15; in addition, 26, 27; in 
subtraction, 82; in multiplica- 
tion, 55, 70-78; in division, 66, 
81. 

Imaginaries, 202; operations with, 
203; graph of, 205. 

Index laws, 47, 58, 59, 187. 

Involution, 56. 


Logarithms, 292. 
Lowest common multiple, 106, 143. 


Monomials, definitions of, 22; addi- 
tion of, 28, 24; subtraction of, 
30; multiplication of, 48, 49; 
division of, 59, 60; H.C.D. of, 
104; L.C. M. of, 107. 

Multiplication, 89-57; of monomi- 
als, 48; of polynomials, 49, 51; 
identities in, 55, 70-73 ; of frac- 
tions, 119, 120. 


Notation, 280, 284. 

Number system, the, iii, 1-3. 

Numbers, incommensurable, 3, 192 ; 
general, 4, 9; literal, 4; oppo- 
site, 10; negative, 11; positive 


INDEX. 


and negative, 11, 12, 18; alge- 
braic, 18; commensurable, 191 ; 
irrational, 198; rational, 200; 
complex, 205; ratio of, 255, 


Ordinates, axis of, 148, 149. 


Parentheses, 7; removal of, 33; in- 
sertion of, 34. 

Pascal’s triangle, 78, 79. 

Perinutations, examples of, 268, 
270 ; definitions of, 269; symbol} 
of, 270; factorial symbol of, 272. 

Polynomials, definition of, 25; ad. 
dition of, 25; subtraction of, 50; 
multiplication of, 49, 51; divi« 
sion of, 60, 61; square of, 73¢ 
H. C.D. of, 105) EG aie 
107 ; square root of, 176; cube 
root of, 182, 184. 

Proportion, definition of, 263; theo 
rems of, 263-265. 


Quadratic, the pure, 208; see Equa . 
tions, 


Radicals, definition of, 191. 

Ratio, definition of, 255; terms of, 
255; compound, 256; laws of, 
256; limit of, 258. 

Root, of equations, 16, 126, 140, 
147, 222; even and odd, 175; of 
decimals, 182; the double, 213; 
irrational, 214 ; complex, 215. 


Signs, algebraic, 5, 6, 138; of aggre. 
gation, 7; in addition, 20, 21; 
in subtraction, 28, 29; in multi. 
plication, 42, 43; in division, 58 ; 
in factoring, 87; in fractions, 


INDEX. | 


110, 111; radical sign, 174; in 
roots, 175. 
Square root, 174, 176, 177, 179-181. 
Substitution, 4, 8; in elimination, 
158. 
Subtraction, 28-38; of fractions, 
117, 118; in elimination, 161. 
Surds, definition of, 192; expressed 
graphically, 192; forms, 193; 
kinds of, 194; quadratic, 194, 
196; polynomial, 197 ; conjugate 
binomial, 198; trinomial, 200; 
and rational numbers, 200 ; 
square root of binomial, 201. 


Trinomial, definition of, 25. 
Type forms, iv; in multiplication, 


51, 57, 70-76; in division, 59, 
81;-in factoring, 85, 86, 88, 92, 
94 ; in linear equation, 137, 147; 
in square root, 176; in cube 
root, 182; in index laws, 187, 
188 ; in quadratic equations, 207, 
208, 209, 211, 216; in permuta- 
tions, 271 ; in combinations, 275, 
276; in arithmetical series, 280, 
281; in geometrical series, 284, 
285. 


Variable, definition of, 15; equation 


in one, 125-146; linear equa- 
tions in two, 147-156 ; quadratic 
equations in a single, 207-228. 


Variation, 260-263. 











oF 


“iptays 


oh a alll allt 


ANSWERS. 


1 


ee Oe 

Page 8. 
24, 4. 22. 130 17. 2. 
29. 8. 39. 13. 20. 18. 52-72, 
Tee 9. 49. 14. 7. 19. 10387. 
151. 10. 49. 15. 24. 20. 193. 
123. 11. 123. 16. 136. 

Page 9. 
(10 b + a) cents. 10. (x + by + cz) dollars. 

: vs years. 11. Va? + b? rods. 

: x , nb 
(6+d) miles. (bc+de) cents. An. & + =a) Hours. 

. © acres. abex 
160 13. 27 cents. 
abe 

fo COUATS, np ah fe 
160 14. et ( ay a |} dollar 

x 2% 
— + — | hours. 
i a 4 3 ) 
Page 17. 

. $1050, A; $450, B. 5. $1200, John ; $900, Henry. 
300. 6. $2200, house ; $1500, lot. 
(aire 7. $120, $240, $360. 

. 60, 160, 220. 15. 428 years. 21. $3500, $2800. 
330. 16. 28. 22. 31,360. 

. 780. 175 60,,.96. 23. 600. 

. $520, A; $780, B. 18. 30. 24. 300. 

mel 00. 19. 54, 27, 18. 25. 60. 

BUNS 20. $4000, $ 400. 

Page 22. 
4b. 3. 0. 5. —8abe. 7. — 2aty. 9. 6V xy. 
— 8 ab. 4. 0. 6. 10 x?. 8. axy. 10. 0. 


fa 


oA FS 


ee 


Si sae ane aed 


THE ESSENTIALS OF ALGEBRA. 


Page 25. 
— 5 xyz. 6. 0. 
5 a2be. 8. «2 + y?. 
Lies, 9. 0. 
— ax. 10. 4(a?—2y). 
Page 26. 
19%—38y+22. 6. 
Tax —5by+4cz. vf 
242+ 4 y? — 3. 8. 


13 a? — y2 4+ 4 2%. 
Page 27. 


Page 29. 
8 x2y. 6. — 16 ay. 11. —15(@+y). 
12 ab. 7) 0) 12. — 4(a? + dB). 
— 21 xyz. 8.10. a 13. 25(ay + yz + 22). 
dg, 9. 29V xy. 14. —17(ax+ by+c). 
hs 10. —4(a+ bd). 15. 5(y? +4 az). 
Page 31. 
—5ai—4474 18 y. 9. 10 xy? + 2 xy? — 3 wy — 2 yf. 
4 at — 25 bt — 10 a?b?. 10. 8(a+ b)—-11@+y)+6ce+5. 
—Sxy + 7x2 + 2 2. 11. 26Va2y —2y3— 162 —17y. 
165 ax — 33 by + 16 ¢. 12. 194ax-+5y?— 23 ez—-l1ly—8. 
—8V2+ 18Vy — 164. 13. 54 y3 — 44 23 + 24 yz. 
—e—e7?4+34411. 14. 36 a2 + 10 a2b — 34 ab? — 12 6. 


ll. Vx+vy. 


— 8ay + 2yz + 18 2x. 
2Vxy — 2V yz —V 2x. 
—6abe+4+ 18 cyz — lmn. 


No answers given, as student is to verify his results. 


15. 1822+ 17 ay +17 yz — 18 y? — 37, 


Page 32. 


No answers given, as student is to verify his results. 


pak ap inact 


Page 33. 
a—4b+4+¢e. 7. Wat+2y. 12. —8., 
x—y-—a. 8. 10x%—18a. 18. 21: 
—20b. 9 6x%—5a. 14. 46. 
—10a+40b. 10. — 25. 15. 22. 


5 ax — 15 by. We eu 


i 


ANSWERS. 3 


Page 35. 
1. 3a2?-—(—4abd+3c-—d). 
2. ay—(— by+ cy + ay). 
38. 5—(d8%-—ar+4abz). 
4. Ty+382—(—24+382x+4 dyn). 
5. 18 — (Sax — 11%+ 16 ax) + 274. 
6. 27 y — d0y — (— 22 zy 4+ 15 bx — 2127) — 80a4 165. 
8. 11 —(8ac—7axr) — (11 by—5y) 4+ 254-344 cu. 
9. b?-—(ab+2ac?—1la)— (4 b*¢y —cy+4y). 
10. 5%+5x2—(2ab+4+6ab+ bax) — (14y — 3 yz — 11 by). 
Page 36. 
(6a+7y—382z)z. 3. (444+ 3y—122)k. 
(8x%+4y—5z)a. 4. (4a—5b)x4+ (6a4120c)y. 
5. (8a—2a)x4+ (40—12b)y; ax —8 by. 
. (44 16a — 12 bd) x2. 9. —2x%+4y+4 8z. 
. 2©—dby4+ 3d. 10, 5 3? 422 4 37 eH 6. 


. (a—a')e?+(b-—)b')y+(c-—c)x+(d-ad)y+e-el. 
. 247417 ay 4+ 4 y?. 
. 644+ (16y4122)k4 (82? + 12) k? + Kk’. 


(a—a@')v?+ (2h—-2h')cy + (64 0!) y? 4+ (ce — c') 2. 


Page 37. 
$400, A; $150, B. 15. (a2—b2)4+3(a+y)—15 V 22+ 92. 
9 ax3+48 bx2y—23 cay? +21 dy? 1%. (1—b)a+(—44+60+4+c¢-—d)z. 
18 22—Ty—6%2. 18. 8%—[—38y4+ (#+y)—2 yz]. 
(a—5+c)+ (3—b—d)y’. 19 ab + cd — LOW Bh al Gad! 
8 a3 + 6 ab — 12 ab? + 6. 100 f 
Zi. 20. $36, John; $72, James; $108, 
462. Henry. 
. 64 yr., father; 18 yr., son. 21. 16274 6ay — 8 y?. 
By — 5 xy. 22. 28 —34%-—27-4y'. 
38a—15ab+76-—11. 23. — 5. 
—2(a+ b) Vaz. 24. 79. 
25. 5(a +y) — 4(a? — b) — 26 Vy? — 
Page 46. 
2ad + 2 bd — 2 cd. 3. 4axr+2 ay —6 az. 
—6ad —9 bd + 12 cd. 4. 48 bx? — 20 by? + 52 bay. 


5. — 30 abry + 26 abyz — 24 abzez. 
6. 5abxy — 25 bey + 15 cay — 5 axy. 


oo 


we ae 2 ge 


ray 


THE ESSENTIALS OF ALGEBRA. 


% —Samxz? + 3 bmx? — 3 cmzy. 
8. 8ablx + 10 abmy — 6 abnz. 
9. 18 a3a3 — 15 aba? + 9 ab2x3 — 3 b328, 
10. — 2 ax’y?z — 2 buy?z — 2 cy?z. 
Page 48. 
. wa, 6. 29.39. 611, 9. (a+b)#(4+y)2, 
ys. 7. q%qltyl8, 10. 79(y+z)4(a—b)}6 
2 ey ttyl, 8. (7+ y)7!. 
Page 49. 
. — 60 ardbx2y. 4. 20 xty%z4. 7. 60 y(a + b)?. 
. 80 a5b2x3y5, 5. — 96 ltxby’. 8. — 80 atz*(a+b)> 
. 40 mnsad, 6. — 6 adic. 
Page 50. 
. 12 a3b — 16 a®d?2. 5. 6 l4xt — 8 [8x5 4 10 arty, 


. 15 akx2y — 10 a2a3y +15 ax?y2, 7. — a'b? + ath? — a3b4* + a?d®. 
8. 3 a5b3 — 4 atb* — 6 a3b® + 7 a?b?. 
9. 6 a®bx? + 6 ab3a? + 6 abet + 6 abaty? + 6 abxrz?. 
10. — 30 abc + 72 be?d + 18 befh. 
11. 20 x3y2z2 — 35 w2y3z? + 15 x2y2z8. 
12. 24 atbh+c? + 56 a2btct — 112 atb2ct. 
14. 5 xy(a? + y?)4 — 15 a2y (a? + y?)3 + 20 xy?(a? + y?)2. 
15. 2x2y(a — b)® — 6 wy(a — b)® + 14 wy?(a — D)t. 
16. — 38 a3b?(~ + y)® — 2 ab?(x% + y)® + 38 a3b4(x + y)?. 
17. — 20(42 + y)® + 12 a3(a? + y)* + 28 ath (a? + y)3. 
18. — 50(a? + 3 b?)7 + 30(a? -+ 3 b2)5 — 60(a? + 3 b?)4. 
19. 5(a+60+c)'—15(a+ 54 6). 
20. — 5(ax? + bx + c)* — 20(ax? + bx + €)3. 


Page 52. 
16 «2 — 9 y?, 8. 22+32— 10. 
a2x2 — b2y2, 9. yt + y? — 12. 
x + Sgt 8. 10. ax? + acx? + bx + be. 
2+ 4 = 6, ll. 12224 262+ 10. 


5a5 — 14 a4 — 10 a3 + 24 a2 —3 a — 18. 
14. 7647 y5 — 8 yt — 22 y3 + 40 y? — 48. 
15. 66+ 464 — 1163+ 138 624+ 46 — 10. 
16. a + 9 at — 48 a? — 38.4? +17a— 30. 


sole os 


ao Femi ipeee gal oh 


ri acct oi ak ado Ae 


ANSWERS. 5 


19s 3 oc + 12 ot + 91 ze? — lida? + 82 2 — 20. 

18. 5 xy — 8 wy5 — 54 wtyt + 35 x3y8 4 52 x2y2 — 18 xy — 12. 
19. 42—102* —17 22+ 8422+ 2-12. 

20. 4a2+ a7’ +2a8—a—8at+7a®—2a24+5a—-—4. 

93. 3a° + 7 a> — 33 at + 39 a? — 29 a2 +19 a —6. 

24. a — 16 a°+ 40 a — 41 at + 76 a? — 74 a2 4 28a — 49. 
25. xoy® — 16 xty* + 48 x3y3 — 76 x2y? + 60 xy — 25. 

36. 6 2' — 925 — 295 + 21 xt — 20 28 — 22 22 4+ 31 x — 20. 
27. 28y7 — 20 y® — 56 y5 + 59 y* — 21 8? — 24 y? + B5y — 4, 
28. — 4 a®b® + 29 atht — 28 a8b? — 30 a2b? + 35 ab. 

29. 4a7b7 — 2 a®b® — 17 atb*t — 13 a3b? + 33 a2b? — 22 ab 4+ 35. 
30; 327° — 275 — 102+ + 14234 42?— 122 — 15. 


Page 56. 


No answers given, as student is to verify his results, 


Page 57. 
16 al2y4, 4, 410 718/24, 7. qidxl4. 9. (a2 + y2)M, 
(a+ y)% 6. 34, 8. 3! . 2472, 10. abc, 
Page 60. 
3 aay. 5. 3a(x+y). 9. 7 lyt(z2—<2). 
3 a2xz. 6. —3(a—b)(a—-y)>. 10. 3?(a — 2). 
BY. 7. —9 x3y, 
3 bed. 8. 38a2z. 
3y — 2 axy? — ax, 3. Bay + 5 y?z —T yz. 
cx’ — 2 abu? + 8 a2c?, 4. («+ y)?+ 5(4+ y)— a. 


5. 4(a—b)—52+4 11 ay(a—b)?. 
6. (e+ y)?—2xy(et+ y)—3y(a+t y)%. 
7. ab +2 a(x? + a?) — 3 b?(xz? + a?). 
8. — «2 —3 yz(a? — b?)%, 9. —2a+38 Bct(ax + b)3. 
10. —3+ 4 a*(ax? + ba + c)3. 


Page 62. 
x3 +2 12 —3 1+-4. 8. bc? — at. 15. @2b?—2 abed+c?d?. 
x8 — 3. 9. m*—2m?+4+1. 16. 3aty* —2 x2y?+1. 
$22—44+1. 10. a” +. b". 17. 2?—6224+1227—8, 
227—52+4+7. ll. a + 6, 18), 1-8 a= x2, 
3 a? — 7. 12. a" + arb" + 5, 19. (a+ b)—4. 
a2 + 2ab + b2. 138. 2? —72+4. 20. (x+y) — 2. 


xt + xa? + at. 14. «27-6249. 21. (a+24)+6. 


THE ESSENTIALS OF ALGEBRA. 


22. (m+n) —11. 
23. (a+b)?+ (a+ b)e4+ 2%. 
24. (w+ y)?+(a+y)(a +b) + (at b)% 





Page 64. 
e+a2?#—4x4—4; c2—22? 8. y6 —8 yt + 64 y? — 512, 
—x+2, 9 at+a*41. 
v8 —602+524+12; 2-32? 10. 4a?-—2a+1. 
—-2£+3. ll. «2? —22+4+2. 
x? —a2— 6. 12. 3b—1. 
2 + 7 y?. 13. a? — 2 a?b + 2 ab? — 6. 
m3 — 7 m?n + 5 mn? — 4 ni. 14. «+y. 
a2 — 16. 15. —2a?+ 8ab. 
644+ 9624 81, 
Page 66. 
x? — 10% +4 41, rem. — 150. 8. 1—1L a+ 72 x — 395 23, rem. 
x2 +44 —10, rem. 30x — 7. 1gy® at, 
a? — 16 a+ 80, rem. — 379. 9. 1—x—8 «2? — 181 23, rem. 
b2 — 23, rem. 218. 2056 at + 655 2°. 
ax? +6 ax — 18, rem. 47. 10. 5+ 22 me aie ay rem. 
14+2%+207+42 48, rem. 2 xt, 1255 4 
1424-22 —143, rem. —i ct. 


Page 67. 


No answers given, as student is to verify his results. 


Page 67. Review Exercises. 
124. 3. 823 — 16 y3 + 23 — 4 ay — ay? 
488. 4. —4—4a4+20643¢e. 
— 16 4? — 31 xy — 21 y”. 
(a+4a4+3l)¢+(b—-4b'4+38m)y4+(c4+8n)24+4e. 


24—-2y—22; —2. Lit: 
3a—25b+421¢; 16. 12. y2z — xy. 
5. 13. 2+ 4241. 
3 V41 —4 V3. 14. «2 + y84+1-—32y. 
3 15. xt — 4 xy? 4+ a3y? — 4 yt, 
36 x7 — 2126 + 98 25 — 88 xt + 79 «3 — 69 a2 + 402 — 16. 
—7; —18; —5. 20. 3a2—a4 + 2. 
Cat cede pedo, 21. a—b—c. 


a*—38a?b+2 ab. 22. «—b. 


ANSWERS. i 


23. 22 —(b+c)x+4 be. 25. atl + quar + ax + antl, 
24. z2—™m. 26. x2" — yn, 
27. 6 a — 12 grt) 4 42n — 20 unt] — yn — 284 — 21. 
28. 3a" 4+ 4a"-14 3 yr”, 
29. om — ym 1. 
30. a? —2 a2b + 2 ab? — DB. 
31. 6x — 12 x74 — 2 xr — opt — arptr + 10 gat, 
32. qn + arbn + h2n, 
833. w2nt2 +- gent] — 10 92" + 13 72n—-1 — 5 92n-2, 
34. x — y”, 88. (Vax+b)?—y(Vax+ b) + 2. 
35. ai” + a2nbin + D8, 39. (x — y)%. 
36. (a+ bd)” — x. 40. a—b. 
37. (a2 + yy) + (a2? + y)" 4+ 1. 
Page ‘71. 
1. 42+8% — 20. 8. 1644+ 8 x? — 35. 
2. 947+ 21446. 9. 94+ 6 xy — 24 xy?. 
3. 1902 —7a—20. 10. (a+ b)? —(a + b)— 30. 
4. 254+ 40a+412 a2, 11. (a+ 6)? -—(a+4 5) — 56. 
§. 4y2?+2y — 42. 12. (x+2ab)?+4(%+2 ab) — 21. 
6. ax? — 5 ax — 66. 13. 20 -—(#+ 2y)—(%+ 2y)?. 
7. 9 0?y? — 33 xy + 30. 14. (a+ b)4+ 3u(a + bd)? — 28 x2. 
15. 15 a7b?—2 ab(x—y)?—(x-y)!. 
1. 27+ 2axy+4+ y?. 5. y2+2y(a+6)+(a+ bd)?. 

2. 407+ 4axr +4 a’. 6. 927+627(2a+c¢)4+(2a+¢)% 
3. 92? + 24 bx + 16 52. 7. (a—38)?4+10y(a — 3) + 26 y?. 
8. (a@+b)?+2(a+d(@+y+(at+y) 

Page ,72. 

1. a? —2ax+ x2. 4. Dat —6 arby + by. 

2. 9a27—6Bay y?. 5. (a+ db)? —2 xy(a + bd) + x2y?. 
3. «?y2 — 8 bry + 16 D2. 6. (8 a+y)?—2 ab(8x+y) +4752. 
7 Ce+y)?—2(@+y)(a+ b)+(a + dD)? 

8. (2x—y)?—6 xy(2u —y)+ 9 xty?, 
Page 73. 
1. x? — y?. 6. v#+ 443+ 4272 — 256. 
2. 9a? — bd. 7 a+2ab+4+ b?— 4? —2 ay — y2. 
3. 16 42 — a2b2. 8. a*xt + 2 abu + bx? — c%. 
5. 4ay + 4 y?. 


OIA Th oD 


es 
o © 


on 69 


THE ESSENTIALS OF ALGEBRA. 


Page 74. 


a+ a+ y2—2axr—2ay+2 xy. 

9 a2 + y2 + D2 —6 ay + 6 ba — 2 by. 
4249624 y2+ 12ab —4 ay — 6 by. 

9 a2 + 25 a2 + 4 y? — 30 aw — 12 ay + 20 xy. 

4+9a2+ 1662+ 12a— 16) — 24 ab. 

et y2+a2+ bh? —-2ay+2axn+2be —2ay —2 by +2 ab. 

m? + n?+ p? + q2?—2mn—2mp —-2mq+2np+2 nq +2 pq. 
4e274+ y2 4922+ a? 4+ 4 ay —12 42+ 4 a0 —6 yz + 2 ay — 6 az. 
a* + ab? ++ 427249 y?—2 0b +4 a*x —6 ay—4 abx+6 aby—12 xy. 
4u2+ yt 4+ o2y2? + a —4 ay? 4+ 4027 —4 an —2 xy3 4+ 2 ay? —2 ary. 


1. w+(c+d-+ e)x? + (ced + de + ec)zx + ede. 

2 @8+(a+yt+ za? + (ay t+ ye + ex)at xyz. 
b3+467+ 6-6. 6. «xy? — 6 xy? — 9 ay + 14. 
y>—2y?—5y4+ 6. 7. 2 — 49 x? + 190. 
m3 — 2m? — 23 m + 60. 8. 7° — 93 y? + 308. 

9. (x+y)? + 10(~+ y)? + 31(4% + y)+ 30. 

10. (8x%+ y)? — 28182 4+ y)+ 48. 

11. (ax + b)? + 5(ax + b)? — 84(ax + b)— 80. 

12. (ax? + bx)? + (ax? + bx)? — 10(ax? + bx) + 8. 


Page 76. 


1. 8a3+ 12a7b + 6 ab? + 6. 3. 27 43 + 10822 + 144% 4 64. 


15. 
16. 


a? — 9 a2b + 27 ab? — 27 b3. 4. 8 «?—60 x2y+ 150 xy? —125 y3 
5. 27 43 + 27 abx? + 9 a®b2x + a3, 
6. 8 mn? — 12 m2n’pq + 6 mnp2q2 — p3q?. 
12628 —7524+15a2—1. » 8. 216, — 1842-0 oe 
9. 64 a3b3 — 240 a2b?y + 800 aby? — 125 y3, 
PE Ory Olek 
1l. (a+ y)? — 6(@ + y)? + 12(4 + y) — 8. 
12. (x—y)?4+ da(a— y)? + 8a2(z — y) + a. 
18. 8(2 + y)? — 36.a(a + y + 54 a2(a + y) — 27 a8. 
14. (ax + 6)? + 3c(ax + b)? + 3c2(ax 4+ b)4 c3. 
(a+b)? + 3(a+ b)*(e+ y+ 38(a+ d)a@+y)*+(@+y)* 
(ax + b)® — 3(ax + b)?(cx + d)+ 3(ax + b)(cx + d)? —(ca + d)3. 





Page 79. 
2. a®& + 6a5d + 15 ath? + 20 a3b3 + 15 a2b4 + 6 abd + 5S, 
3. at + 7 axoy + 21 wy? + 85 xty? + 35 x3yt 4+ 21 x?2y5 4+ Tay? + y7. 


ANSWERS. 9 


m8 + 8 mn + 28 mon? + 56 mon? + 70 mint + 56 m3n5 + 28 m?2ns 
+8mn' + n8, 


. 14924 867? + 84273 4+ 126 x4 + 12625 4+ 84278 4+ 386274 9 784 2, 


y + 10 79 + 45 8 + 120 y7 + 210 95 4+ 252 y5 + 210 yt + 120 73 
+ 45 y?+ 10y 4 1.- 


. 162! + 82 ax? + 24 a?2x? + 8 a3x 4+ at. 

. p’—8 p'q+28 p'g?—56 pq? + 70 p4q* — 56 p2q° + 28 p?g® —8 pg? + q°. 
. Bl yt — 216 y3x + 216 y22? — 96 yx? + 16 x4. 

. 824° + 80 atd + 80 a3b? + 40 a2d3 + 10 abt + DS. 


© — 15 a4y + 90 x3 y? — 270 xy? + 405 ayt — 243 y?, 
(21 x)® + 6(21 x)°a + 15(21 &) 4a? + 20(21 w)3a3 + 15(21 x) 24 
+ 6(21a)a° + a®. 
xvl2 — 12 xy + 60 x8y? — 160 xy? + 240 xty? — 192 x2y5 4+ 64 7/8. 
— 9 xl6y2 4 36 altyt — 84 wl2y6 + 126 xly8 — 126 x8yl) 4+ 84 8712 
— 36 xtyl4 + 9 x2yl6 — 18, 


» + 4aly + Tady? + 7 ary? + 3 axtyt + ey? + ye wy? + ys ay" 


+ ghz ys 


. dx? — 7-38. 4 aby 4 21-35. 42x°y? — 35. 34. 48aty3 4 35. 38. 4434 


— 21-32. 45a7y5 + 7-3 . 46ry6 — 47y7, 
(a + y)* + 12(4 + y)3 + 54(a + y)? + 108(% + y) 4+ 81. 
(a+ b)*—82a(a+ b)? + 2422(a + b)? — 3223(a + Db) + 1624. 


. 245 4° — 405 at(a — y) + 270 a (a — y)? — 90 a2 (a — y)8 


+ 1l5a(e — y)* —(% — y)?. 


ie 4. 5 y+5 oer 32 xy? | 16a2y* 64ay? | 64 y8 
55 5S 52. 33 5.33 5-34 36 
. 2+8r7415. 4 a2x?+ 8ar4+ 15. 7. of — 5 27*— 50. 


v24+8x%—20. 5. x2y2+(a—b)xy—ab. 8. (w4+a)*—4(x7+a)—21. 
v+3e2—40. 6. 1622—122—10. 9. 12 +7 by + by?. 


(x?+5 27)?—3(x?4+3 2) —10. 15. 427 — 16 y?. 

. 92? — 3xy — 2 y?. 16. (2% +4 y)? — 64. 
(a + b)? + 6(a + b)— 72. 17. (a? + y?)? — 16. 
(x ~5)2— 18. (ax + by)? — c?. 


19. 2* + 27+ 1. 


. 9x? — y?, 20. (xy + yz + 2x)? — a2, 


21. 224+ 4724+ 224 2xey+2yz24+ 2 eu. 

22. 902+ 4y24 224 1l2ay4+ 6924+ 4 yz.. 
23. a2x?2 + b?y2 + c2 + 2 abay + 2 ace + 2 bey. 
24. x27—-6x4+ 9. 

25. 474+ 9y?+16+4 82+ 24y + Bay. 


10 


wo © © 29 
woes © 


Sen oe a ea 


THE ESSENTIALS OF ALGEBRA. 


26. 942+ 4y?4+ 36 — 36a — 24y 4 122y. 
Dra gn — 2 ynynr + yn, 
28. x2 4+ yy? +224 w?+ 2ry+ 202+ 2aw + 2yz2+ 2yw + 2ew. 











. 2+ 14424 638% +4 90. 33. «3 4+ 93, 
et — 3427 + 225, 34. a? 4+ 6? 4+ 3 — 8 abe. 

. 02 + 27 a8. 35. 2+ 8y3+ 1-—62y. 

» 8+ 9x7y + 27 xy? + 27 y?. 

Page 81. 
a— 9. » 18 eR 
442 + 62, 15. y"+ 8. 
3+ a. 16. a+b—3%. 
4a?—2ab+ 6. 17. vw" — 5y", 
9a — 5 db? 18. («+ y)"+(a+ db), 
1— da. 19. av+b+ 52. 
y — 2. 20. ax? + b4 +¢— lx —m. 

. (4+ b)2+4(a+ b)+ 16. 21. y2” 4 yranr + 12”, 
9a2?+3a(e—y)+(e—y)% = 22. _-¥yBr 4 yPrgnr 4 yrg2m 4 xn, 
(1+ y)—2ab. 238. V3z+y4+ 4. 

ny a? 24. (Var+by+4yVar+b4+16y2 
epee Lbs 25. (Vaxr-+by)? —5a2Vax + by+ 

. ©+y—12(a+5b). 25 at. 

Page 84 
a(a + y). 15. axy(l+at+say). 
3 ax(1+2 ax). 16. 38axy(a—54+4 7 ay?). 

. 5a(a4+264 be). 17. S5axyz(a? + 6 xy — 8 2). 

. ey(d5 —Caxry +7 xy?). 18. 7 a®xyz(x? + y? + 27). 

. m“(3+4mxr4+2y). 19. 3abc(aVe+y —2bVx%—y). 
21(—24+2Wy+4+38y). 20. 5ayz(aVar+b+2yVax—b). 

Page 85. 
(a+2b)(a—2b). 5. (24+82)(2-—382). 


(2a+3b)(2a—38b). 6. (day + 62w)(S5ay —6 zw). 
(ab + c)(ab — Cc). 7 («tytaj)(x+y—a). 
(4ab+5c)(4ab—5c). 8. («+ 3y+2)(4+3y—2). 
(8%—2y+z2+a)(8x—2y—z2-a4). 
[2(@ + 2y)+ 3(a + b)][2(# +2y)— 38(a + d)]. 

[2”%+ 4(x% — 34) ][2 a — 4(x — 84)]. 

(4 + a*)(24+ a)(2— a). 16. (8%+4y+4+8)(82+4y4 2). 
(x2 + 4 xy + y?) (xa? + y?). 16. (—2%+38y—5)(—4x2+y—8S). 


a 


ANSWERS. 11 


(lx + my+n-+2 an +2 by+2c)(le+my+n—2 ax —2 by —2¢). 
(ax + by + 2lz+2 mw)(ac + by — 2 lz—2 mw). 


~ (arn + Ory”) (Caran — Bry”), 21. («#+a+1)(e«—a+1). 

~ (ary 4+ Zw") (ary” — 2"), = 22. (44+ 2y4+38)(e-—2y438). 
. (84+38b+a)\8x—8b4 a). 

. [(@+a)?+(a+ b)?7]@+2a+b0)(4—b). 

. 8a+yy-—*). 


Page 87. 

. («© +1)(@+42). 15. (644+ 3y+2)6%+3y+8). 
(x —1)(@ — 2). 16. (x2 + y? — 9) (a? + y? + 8). 
(a+2)(a—1). 17. (2x+38y+4+8)(2044+3y+41)). 
(* —2)(%+1). 18. («+1)(@+2)(@+8)(4%+4 4). 
(ax + 2)(ax + 3). 19. (7©+38)(%@—38)(“#+2)(@ — 2). 

-. (@+2y)(*4#+3y). 20. (ax + by +7) (ax + by + 1). 
(xy — 5) (xy + 2). 21. (ax + by —1)(ax + by — m). 

. (1+ 5ay)(1 —2 ay). 22. (1+ 38)#(a + 2)(a + 4). 

. B«a-y)2Qae-y). 23. (427-— 54+ 2)(a7-54+4 5). 

. (©+y¥4+4)(4+y4+ 5). 24. (x — 8)(a" — 2). 

. (@+3860—5)(a+3b+44). 25. [(ax)" + I][(ax)" +m]. 

Page 89. 
(x — 4). 15. (a? 4+ y? — 2?)?. 
(2% — 3)? : 16. (x+y+2z)?. 

- (ax + 5y)?. 17. (a—b—c)? or (—a+6+40c)?, 
(7 ab — 1)?. 18. (244+ 3y+4 2)? 

. (10 — ab)?. 19. (a + 6)?. 

. (ae +04 c)?. 20. (y*” — 7)?. 

. (8a+4y—3)2. 21. (a" + 6 b*)?2, 

. (22-38 y)?. 22. [((a@+y)"—saj 

. (ax + by+e4 4)? 

Page 90. 


4. («© —2b)(a? + 2 bx + 4 0?). 
5. (2a—38b))(4a2+6ab+4 9b). 
6. (ax + 4)(a2x? —4ax+4 16). 
7. (ax —3 yz) (a2? + 3 axyz + 9 y?z?). 
8. («+y—S5z)[(e@t+y)?+52(e+y) + 25 27]. 
9. (84+442y)[(82+4)2?-2y(82+4)+4y?]. 
10. «(a+ 3)[(22?+ 3844 4)?4+4(@?4 82+ 4) 4 16]. 
11. (ax + by — cz)[(ax + by)? + cz (ax + by) + c?z?). 
12. 5@+y)[Gx+4y)?—-Be+4y)(2e+y)+ 2@et+y)?). 


12 THE ESSENTIALS OF ALGEBRA. 


15. (Qa — b?)(4 a2 4 2 ab? 4- D). 

16. (ax? + by?) (atx* — a?b2a?y? + bty*). 

17. (w—1)(#4+ 1) (a? —a44+ 1) (2? 4+ 441). 

18. [a%x? + (y+ z)*][atat — ata? (y + 2)? + (y + 2)4]. 

19. [(ax + by)? — c*z*][(ax + by)* + cz? (ax + by)? + ctz4]. 
20. (4 a? + bc?) (16 at — 4 a?b?c? + dtc). 

21. 4 xy (8 a? + y?) (a? + 3 y?). 

22. [(a+ b)?+ c][(a+b)*—c(a+ bd)? 4 c4]. | 














Page 93. 
1 (8a+y)(e*+2y). 5. (ax +3b)(6x44+ 5). 
2. (84+2y)(2x—33y). 6. 25y—32)(2y—~2). 
3. (7xe+y)(8x%+4y). 7% 2(4y+2)8y—42). 
4. 2sx—y)(2x2+y). 
Page 97. 
1) Ba ed al) (eee 
ore. 89 2 oat 
9 —2(a— 24 aR) - 7-3). 
2 2 2 2 
3 (2-245) (eB), 
6 6 6 6 


4 1 Cae ee (x-38-¥=*). 
2 2 aie? 2 


eee (2 rele xe (« 4" sp a 











2 2 2 
6. 11(2—34+ YH) (2-8 - v 
2 2 2 2 
7. —4(x%+8)(x—1). 
2 2 2 £ 
9. -8(2 84 21) (2-332) 
2 2 2 
10. 10(2 244) (x -7_ A), 
2 2 2 2 
Page 98. 


l. (w@+a4+3b)(4+2a+4 Db). 38. (x —2y+32)% 
2. (t+3y+b)\(2e4+y-—D). 4. (2a+3b—4c)% 
5. (2a4-—S38y+z)(a@+y—382). 


ge alae ool ea! 


a 


ll Dg 


a a 
OP WON OO SAD AP ww Pr 


- (@—2)(a+1)(a—-1). 


ime Sas Sk 


ANSWERS. 


Page 99. 
8. (4a+6)(8a—b+40). 
4. (x+4a)(4—5a+y). 


(2%—-—5y)\(2x+7T7y-— a). 
(x—4y)(e+3y+4+2). 


13 


5. @aty@r—y tp). 


Page 101. 
6. (x+2)(@ — 5) (x + 8). 
7 (© +2)(%+ 8)(@4 4). 
8. (7x—1)\(@+ 1)(~+4+ 15). 
9. (x+3)(%+4)(~+ 5). 
10. (7—4)(a—4)(x—4) or (x—4)3. 


(7 —1)(% — 2)(%4 2). 
(w — 2)(% + 2)(x% 4+ 2). 
(«© +1)(%—38)(#+4 5). 
(% —1)(% — 2)(a% — 8). 
(2 — 2)(% — 8)(a% — 5). 


Page 102. 
b(a@ + 1)(#— 1). 16. 
10c(m + 2c)(m — 2¢). VG: 
(x+z2)(~1+y). 18. 
(m—r)(l+ 7). 19. 
(7 —2z)(2x+4 3y). 20 
(38 — a)(2— a). 21 
(x? + 9 y*) (a? — 8 y?), 22. 
(m — 12)(m + 8). 23 
a(a + 6)?. 24 
(8% +1)(%+4+1). 25. 


2(7+4(m+1)][7—4(m+1)]. 926. 
(8 a? + Dy?) (8 a? — Dy?). 
(8 —4%7)(9 + 12474 16274). = 27. 


e(4y+nr)(4y—~2). 


28. 
29. 


(m? — 2 n2/?)2, 
4a(b+¢). 

(7a — 4)(2a? — 3). 
Rwy Ei Ci 0 ioe 1). 


. (84%Ay7¥4+82)(54—-—y-—382Z). 
. 9(a — 2b)?. 


(a + 5)(a + 24). 


. (77+ 11)(% + 8)(% — 38). 
. 0(2141)(414+1). 


(% — y)(a — 2) (a — 1). 


Beeb 4% 
ate )(s *) 
(m?+1)(m?+m+1)(m—1). 


a*b*c?(1 — ac)?. 
(a—y+1)(e—y-1). 


30. (1662+1)(2b4+1)(2b—1). 


Page 102. Review. 


x—-5y. 10. 
0. 1) 
6x—l6y+4z2, or —14x%4 12. 
10 y — 62. 13. 
a’ — 16. 14. 
x" — 2ary + y?. 15. 
(a + bx) (x+1)(@—-1). 
2 ax — x?. 16. 
x2 + 2ay + 2 y 17. 


18, 35. 


08 + 2 x2y +2 xy? + y, 

(a —1)(%+1)(x—8)(x + 3). 

(a8 + ab —ac—y?+ 11)z. 

— 421. 

54, 

4q64+9 22 + 25 —12 a3x+20 a3 
— 30%. 

gmt l _ yrtl, 

q2m+l1p2 4 ambnts + qmt4pn-1 + 
azbn, 


14 


18. 
20. 
21. 


dha on pe 


phate bee el er 


— 


a mE 5g Pek al li ecole 


THE ESSENTIALS OF ALGEBRA. 


ain + 2q38 —2aqr—1. 


0. 


22. 
23. 


5a(y—z)™-4*; Sa(y—z)m-*; 24. 


4 x2 (y—z)4m-8 : 4, 


4 yz. 
5 a?b, 


, Ar ee. 


x— Y. 
a+ b. 
x+y. 
x— 3. 
a+. 


12 a®baty. 


plo alr, 


24 p2q3r2, 


a(a — b)*(a + Db). 


(7 — 4)(% + 4)(@ — 5). 
(p — 5)(p + 5)(p + 6). 
(LEB) 0) 79: 


25. 


(a+ 1)(a—2)(a — 38). 
—y?—ye— 22-2 xy—447242 x2, 
4(8a2—19y). 

b@—jath. 


Page 105. 


4. 7 ab?m3. 


7. 3(x — y)32°, 


5. 22 p2q2x3y. 8. 9(a? — b?)2xy. 
6. 2(a — b)?x?y. 
Page 106. 
6. b—7. 11. s4(4-F Oe 
7. «© — 2. 12. (x+y). 
8. b. 13. m — 10. 
9. x2(4—y). 14. x—y. 
10. x3(% — 3). 15. a—2. 
Page 107. 


4. 150 l*m3n’3. 
5. 21 a®bta ry’. 


a3(a —2b)(a +2). 


6 a?(a? + 3)(a? — 38). 
(m — 1)(m? + m+ 1). 
(r —2)(r—- 8)(r+4+ 8). 


a 


dbx 
ay 














7. 20(@ — y)8a3b3. 
8. 120(a? — x)4y?z2. 
492, 


e(1+42z)(1-—472). 

(y —2)(y —TD(y + 8). 

(a 4+- 8 b®)(a — 2b). 

(ec —3d)(?+3cd+ 9d?) 
(c+2d). 

(16 — zw). 

(8 + b)(8 — b)(9+ 3b + B?). 

(x — 3)?(% + 3)2 





x 


6. 12(a — b)2x 
Page 108. 
9. 
10. 
11. 
12. 
13. 
14. 
15. 
Page 113. 
Bye 
b 
g 
b 
D etLD: 
Sie 
r—a 


9 @teyty, 


a+y 


. Nocom. factors. 
1 


+1 


15. 


16. 


ANSWERS. 15 









































1 ; 18. —1. 91. 2% —Yy. 
Oy +P 19, 242. ny 
Sha Ale x—l 99. Tty—%, 
x? — xy + y? 90, 2=2. xe+ty+z2 
ee ay + yt x + 4 grat 
(x+y)? x 
hath pe vai Tif jens nL SSR 
%—C xr—y (1 —a)(1 — dD) 
%— 3 28. «x. 31. ie UEC, 
x—5d 3 a+b—ce 
x+a 
r+e 
Page 115. 
x+1. 7. x2 —Qary 4+ y?. 
width, Gh fe 10 eo Gy ee 
9 ev2+ae+1 
ee a TON en ee oe 
a2 + 3445 ages 
% — 24+ ——_—_—_.. , 48 
e+ 24+ x Il. 1+ 22 4+ ot + 2h + : 
e+ 3a. 1 — x? 
gz 9 ome De 
Jaypee 12. Be ea ae a a 
Page 117. 
3bc 6ac 2be bac 3ab_ 
3abe Babe Babe Babe Babe 
yaty) (e+ y) Yee: 
cy(et+y) ey(at+y) xy(et+y) 
3(z@—y)i 4(e+y). 
gz — 2’ 2 — y? 
a(x + 1)(z + 2) ba(xe + 2) co{e+1) 


a(x +1)(2+2)’ x(x+1)(@4+2)’ x(x4+1)(x +2) 
a(a—1)(a2—-4) b(a+1)(a2?—4) c(a?—1)(a@—2) d(a?—1)(a+2) 
(2-1) (a?—4)’ (a?—1)(a?—4)’ (4?—1)(a?—4)’ (#?—1)(a?—-4) 
Be eee -—-6e ba + 10% 


’ 





g?—4’ g2—4 2 —4 

un? —5x 322—6%2 , 
(9) 4) (2 — 5)’ “(4 — 2) (@ — 4)(2 — 5) 

e-+-1 e+2 2+3 f 
(%+2)(%+3)(x+1)’ (%+2) (a wee eG (+2) (*#+38)(*+1) 
ax? + axy? bx ca? + cy? 


x(at — ys)" x(at — yt) x(a — yt) 


10. 


12. 


14. 


16. 


THE ESSENTIALS OF ALGEBRA. 


xy — 4 xy? x2y? + ry? — 2 yt xt — x8y — 2 xy? 
(a? — 7) (a? — 497)’ (7 — YP) @?—4y)’ @—Y)\@—4~) 


14-2442 2742238424 32-323 52 1—2%4+22?— 208+ 2* 





’ ’ 


1—< 1a7t 6 leat 1—z4 
5 2+20 64+18 7x“2+14 ; 
(a+2)(a+8)(a+4)’ (%+2)(%+38)(2+4)’ (2+2)(@+8)(a44) 
tee 2x”7—2ay+2y? 3 i 
+ oy? 4 y?? at + oy? + y4 > ot 4 a2y2 + yt 

abe ae 15 5(a + b) O ame 

prtlem+1? patlem+l : a"(a + b)*’ a"(a + b)4 
Page 118. 

re tk 9. 322+ 3% —3 3. 50%. 4. 2 ; 

a—l x-+1 z+ 10 x+2 
8 + x 6. + 3a +3a% + a2+%—-—a ”. 8 
x—2 (x +a)? a 
ee 10 ee 19. 
z+a 2 —1 (2—1)(a+1)2 
2x7*—62%+3 wu, (2?+6, 13. a? — bab — 6°) 
x? + a8 at — 4 a* — b? 

227+182+4 16 : 17. 3827+182+6 ’ 
(a — 2) (« — 8)(@ + 8) C+) + Date 
Maa taa ta ac IP 18. —2y ‘ 

et —247+1 (xty+z2)(x+y—zZ)(@—y— 2 
__ 40(5%+4) 19, — wl Eee 

(x + 1)?2(@ — 1)? (x — 2)(a — 8) (a — 4) 

11 —32 ; 90. ry + y2+ 2x — 2 — y? — 
(x —1)(@—2)@ +4) (x—y)(y —2)(2-”) 
, Hage 121. 





























cy 10 (ae Oa 16. &+ 3. 

ety  b2 hage ee 

b(% — y) | 1 b2c7d? Lik 

a?(x + y) "  @6 1890: 

7 by*(a?—ay + y?). 12. a? 4904 y* 19. «+1. 

4 ax y? x 90 (4-2) Ge Ty 
13. @ 4 or (y™ + 1)(a" + 2) 











5 


g 


Ne) 
o 
o 
co 
Q 
Cy 


ry 
S 


10. 


(OAD TP © 0 


OIA AP WN 








ANSWERS. 17 











Page 123. 
y. 1 1 a 
ar ; 3. ai 4. qo yt uz ye 5. b 
10 by 
2 x? — 16 xy + 327? oe LE en ap a 
' Oat — 30 x3y + 25 xy? x2(8% — 5 y)? 
Reem re to. 1° ion 1. 1 19.4, 
(x+y)? : “—y 
Page 124, 
Le Bak 10 eerie 
y g, 2(@— 8), + day + 2y? 
£-+ a. aid 1, 10%=3, 
22-24 4 2ax_. dxz—1 
art " gt 4 @? 19, 2*—2, 
e2+1 8. x—y. 2xz-1 
(a+ y)? 9. b+4 
y b+2 
Page 1381. 
8 i Eos 21. 4,7. 32. a+. 
5 12. 10. 22. 3. 33. a— b. 
9 13. 0. 23. 7. 34. c43. 
10 14. 3. 24. 13. 35. 3m+15n. 
3 15. 41. 25. 17. 36. ab(b+ a). 
13 16. 43. 26. 5. 37. b? — a’, 
11. i owe & ry Ee oe 38. — 12, 
41, 18. 34. 29. —(a+)). 39. — 14. 
13. 19. —6. 80. a2?+ab+b% 40. 4. 
10}. 20. 5 31. a?—ab+ b 
Page 133. 
$190, A; $60, B. 11. 130. 
70, 11. 12. 75, 76. 
36, 21, 57. 13. 100, 101. 
$72, A; $56, B; $48, C. 14. 80, 81, 82. 
70, 150, 220. 16. 50 years, man ; 45 years, wife. 
541, 320. 17. 35 years, father ; 11 years, son. 
100. 18. $6300. 
120. 20. $120. 


18 


21. 
22. 
23. 
24. 
25. 


26. 
27. 
28. 
29. 


G0 et 63 Se 8S 


os Ne Pn cone tag 


THE ESSENTIALS OF ALGEBRA. 


$ 60,000 30. Cc ac ; abe. 
37. l+a+ab 14+a+ab 1+a+ab 
iA 31. $800, A; $400, B; $250, C. 
$ 38200, $2800. 32. 20 miles. 

$80, A; $70, B; $110, C; 33. 80 acres. 

$500, D. 34. 5 p.m., 270 miles. 
12 dimes, 24 quarters. 35. 1 p.m., 60 miles from A. 
12, 10 miles by carriage. 





10 cows, 20 calves, 40 sheep. 36. 160 miles by boat. 

120, 121, 122. | 520 miles by train. 
Page 141. 

8, — 5. 15. 2, —5 28. 0, —2, —3. 

—2, —4 16. —3, —5 29. 0, 3, 9. 

—2,—8 17. 8, 3. 30. 4, — 3. 

Sat i 18. —a+b, —a-—b. 31. —1}3. 

412, 1926557, 32. 4, —6. 

2, 2. 20. 1, 20. 33. 3, — 3. 

1, —1. 21. — 3, — 25. 34 6+ Ce ue. 

bibs py Rane bisa. ie a 

5, — 5. 23. — 14, 7. 35. a+b, —(a+b) 

4, —4, 24. 1, —4. 36. 2, —2, 3, —3 

a—b,-—a-+ob. 25. —2, —7. 37. 3, — qb. 

3, — i. 26. —1, 2, — 4, 5. 38. 0, 0, 1. 

2, 5. 27. —1, —2,—5,2. 39. 0,0, §, —4. 
40.01, 1.12 
Page 144. 

3, —3. 7. 41, — 28. 15. 44 hours. 

1,-—1 9. 4,5; —5, —4. 16. 8+ hours. 

4, WR temas $ Lis 2.0re 

5. 11. 9,10; —9, —10. 18. ab days. 

14. 13 Si met MI a+b 

ai hav oe 4 

2 Bde AL tOuper OF (Factoring by Art. 79.) 20. 2. 
Page 159. 

(8, 1) 4. (3, 2). 7. (4:5, 195). 9. (6, 10). 

(4, 1) a Ca 8. (144, 14). 10. (6, 15). 

1,4 6. (—5, 0). 











ANSWERS. 19 
Page 160. 
1 (2) = 1). 5. (129, 214). 8. (—}, 14). 
2. (17, #4). 6. (* a+b rea), 9. (%, h). 
Soe (85 0): 3 3 Tle (0 m)s 
4. (49, 2). 7. (4, 0). 
Page 165. 
Lie yds By (2, 10): Sa (4e1). 
2. (4, — 4). 6. (5, 2). 9. (#8, — 29). 
3. (— #, 2). 7 ae lc—ap\ 10. (4, 1). 
4. (48,, 234). ' \am—odl ee ae ae 
12. Ga bm km + ae 16. ( bq ne ag i: 
In+km ak—lUb mb+an mb+ an 
13. & — bP b? — a? ie 17. (e + mn + nt — mt). 
ac—b* b(c—a) m+n m+n 
14. (2n—m,2m—N). 18. (5, — 4). 
15. (7, 8). 19 (= ="). 
20. (2, 5). “\ab- a — > 
Page 166. 
2. 2 cents, 3 cents. 20. $24, A; $36, B. 
3. 80 cents, 45 cents. 21. 19 or 91. 
_ 4. 36: 22. 10 dollars, 40 quarters, $20. 
5. 7% 23. 200 boys, 300 girls. 
6. 12, 63. 24. 5miles, A’s rate; 4 miles, B’s 
weace, 61. rate. 
8. 30 sheep, 12 calves. 25. 108, 45. 
9. 60 years, 20 years. SG 118.13. 
10. $540, A ; $360, B. 27. 24 feet, 16 feet. 
11. 36, 54. 28. $760, A; $920, B. 
12. 20, 50. 29. $800, 4%. 
13. $180, $330. 30. 27, 19. 
14. 14, 10. 31. 40 men, 20 boys. 
15. 25, 36. 32. i. 
Lig. 33. 40 years, 15 years. 
18. 49. 34. 80 feet, 100 feet. 
19. 12 persons, $10 each. 35. 15 miles, 3164 feet. 
Page 172. 
ae. 10.70, 1). 3. (1, —1, 0). 5. (4, t, 4). 
2. .(12,-8, 10). 4. (14, 10, —8). 6. (G4, 4, 4)- 


20) THE ESSENTIALS OF ALGEBRA. 


mr nr ir 9. (6, 4, 3). 12. 18, 50, 44. 
( 2” a 10. (1, 2, 3, 4). 13. 50, 54, 72. 
(22 24 2n 11. $300, A; $300, 14. 629 
2 ight Le B; $150, C. 


15. 15 days, A; 10 days, B; 30 days, C. 

16. 547 ininutes, A; 80 minutes, B; 40 minutes, C. 
17. 24, 38, 51. 

18. 20, 40, 120. 

19. $500, A; $800, B; $1200, C. 


Page 176. 
3. +: 4 a3bc?. 11. + 25 a'pl5c, 16 9 x2yt 
4. +13 xyz. 12. 10 abc’. ~ 16 a'b2e3 
5. 12 xy". 13. + 20 ad8c9, 17. +2 atb®c?. 
8. +(a@+32). 144 12 ab. 18. — 2 a>xoy. 
Oe Date, Ni eaey 19. + a2bic?, 
io. + &. 15. — 7 ade. 20. — a%bic8, 
cd 
Page 177. 
1 +(@+ 8). §. +(42%+4+7y). 9 +(a—6+4+3¢). 
2. +(8%+4 4). 6. + (ax + by). 10. +(2%—y+32). 
3. +(@-—9). 7% t(e+ty+2). ll. +(a+b+c¢c+d). 
4. +(%—5y). 8. +(%+3y+4+2). 12. +(~7—38y+4). 
Page 179. 
(The double sign is not written, but is understood.) 
eae b, 5. 202+ xy + y?. 9. 2+ 3a ae 
Ais ais 2 2 2 
eee 2x +3. 6. 32 5 ary +2 y?. 10. Y + by +2. 
3. 2274+ 327 +4. 7 l—-ax—8y. 3 2 
4. B+60e7+5x44+1. 8 24-—8y+42. 
Page 180. 
1. 14 boat t dat . 1 oo 
221 + 2a 2 at A ge, 2 
3. l—4xr—f2?- 223, if aa 2 2 4+ fF 
4. 14+4%4+4 327 — 3 2. 3 92 a 
a E x + —— g 
5. 1+ 3% — 4b a? + 195 23. +5 nF 8 "1G 


Page 182. Section 125. 


1. 324. 3. 239. 5. 485. 
2. 528. 4. 248. 6. 1045. 


ANSWERS. 


Section 126. 


21 








4.28. 8. .865. 5. 1.054. 4. 2.236: 
2.94. 4. .0548. 6. 1.414. 8. 3.316. 
Page 184. 
x + 2. 4. a+b+1. % 227—32+ 5. 
e2toa+i. §. a@+6 42. 8. 2a7b? —8ab+2. 
x? — xy + y?. 6. 3a—4y. 
Page 186. 
Lis 3. 91. 5. 138. cp RUE 9. .2008. 
67. 4. 85. Gi 12: 8. 1.442. 10. 1.270. 
Page 189. 
at y* 10.7 1; 
cs ales 1 
4 b6 11. = 
5x 10% gre: % 
2 yes BE 
62 b3 
1 ee Be lat | 
al wa oy SW 14.0% = 
a*b2xy®, 9. ai2p%cl2 y2 yt 
eee Q1. a 3b 26348, 
y? eames $44 25 
x? — y~?, Soir airUr Ce 
3 2-2 __ 7-8 
= a e rae 23. 4 ch + 4 aqbe? 
et—3 4344 2?—6243-—2271. 24. 323y3 —4ab3 


y? 
Dm +- e- 2-2 + yt. 


a343 a-2b-143 a-1b-24 b-3, 25. 
Page 195. 
4V3. PEN Sep 3y/2, 4. 7V3. 


(3 ay? — 2 ay + 13 y?) Va. 6. (5ab — 9 4+ 12 a2b?) Vab. 
(ho GP ESA ee 


8. [ax2(y +2) —38a% + 12(y4 2)]Vy +z. 





V5. 16. «v3 2. 
av2 x. 16. (4a+6b)V3a. 
(m —38n)V2m. 17. (a.— 6 b)V5b. 
—(4%+6y)Vv3. 18. 26V11. 
. (2242y)Vx + y. 19. 48V13. 
. avd +3. 20. (11 a2 — 1802) Vx 4+ y. 


Sree awe 


10. 


11. 


THE ESSENTIALS OF ALGEBRA. 

















Page 196. 
40. | 7. 6(a— b)*Va—0. 
400 V2. 8. 10(@ —y)2(a@4+ y) V3(a3— 9). 
900V5. 9 BV2(¢ +9). 
48 ax? Vat 2V5(« — y) 
(@ —y)?Va —y. 10. 4. 
540 wty?. 
Page 197, 
—2 2-1 8 -1 4 11 5.2 6. 56—12V3. 
7 +2V10. i 12. 42. 
6 + 2V6 — 2V2 —2V3. 13. 8V3 +1. 
84+5V3, _ 14. a—b—c+2Vvbe. 
—14—8v3. 15. 4% —10Vxz + 252 —-9y. 
BV 8 bVb 1116 ay 16. 
Page 199. 
v10. 4 2V2—-2. y —12V6 + 9v42 
2 gecvaal aa 47 
Ee pynnucriie US 8. —(3 + 2V2). 
v 
eres 6. 5V2 4+ 4V3. 9, oe ss 
74+3V5. 19, 6V8—8V15 + 3V2—4V10. 
2 105 
2V5 + 2V8 + VI5 +38. 13. ve oe ALLIS ft 
( 
14, 9V74 6V14 + 8V21 — 6V5.— 4V10 — 2V15, 
43 
15, 2Va—3Vb—4Vab + 4a, 
; a—b 
16, 2Ve—5Vab —4Vb —2b +34, 
: 2a—8b 
17. 6V~a—1+4+94%-8 
10 —92 
ig, @tyve+(et+y)Vy —2v2y —yv2 0, 
x? + y? 


19, tVe+ yVvy. 
%—y 
90. - (24 —3b+4+2V2@—3 ab + ey) 
(a+ b) 





ae! m oN 


rr 
ee ee OOS 


i 


ANSWERS. eS: 


Page 200. 


. 14+Vv2—v3) v2. 8. (V8 + V2 —V5) V6. 

Dewi 2 V7 )(1 —2°V 15). 4.0 (V6 + V2 11) (1. V2); 
HCV 10i4--V2 — V3)(6 — 2 V6). 

6. (Va—Vb—c)(a— b— c+ 2c Vb). 

%. (142 V2 —3 V3)(6-V6 + 17). 

8. (2Va+V2b—38vc)(4a—2b —9ce—6 V2 bc). 





Page 202. 
V3 4 V7. 5. V11+V5. 9. 243 V5. 
vi — v5. 6. Va—vb. LOSE yt 
V3 + V8. 7. V10—v2. ll. 8V7—2 V6. 
V5 — v2. 8. -V74 V6. 12. Vat+b+Vv6e. 
Page 204. 
LO V24. g 243 V5i, 190, Va +2 abi, 
. —15 V2 —10. . ae ae We 
13 —8 V3i. ee ee eV t, Livia 2. 
—~V64+3 V2 7 1, AO +4b45, 
+(24+3V3)i 8 V3. aa ayers. 
. 8+v6i. 9. y> — a3, 
Page 210. 
+6 ih AES v651 
Page 212 
8, —1. 12. 43, 7. 23. —bive?—c. 
3, —3. 13. —4, — 42. 24. m, n. 
3 il. 14. 3 =I. 25. oe —1. 
7 15. 2, 6. a—b 
oy) — 3 16. — 3, — 3 Sap re bere, 
9, —4. 17. — 3, —%. MCRL Lars 
Bae 18. 3, — 1A, 27. 3, Fe 1. 
See € 19 2ee 28. 3, 2. 
ea 20. 2, 6. 29. 3, 5. 
4, — 5, 215.110; 99, poe VE +4 ab, 
4,—3 22. +6. 2a 


i" 


ve 


a i 


THE ESSENTIALS OF ALGEBRA. 











Page 214. 

Bio 3. —4,-h.) 5) —}, 1. Fee 9..° +8) 8: 
L$) 48 6,402: —3, —%. 10. 3, —4. 
Page 215, 

Bg bs BAVA if 4 3 + V69_ 7, S£VO6T. 
8 \ipaed 0 d 4 
T4+VB ee aut eee 
Buen ; 5 g, 43+ 11299. 
a 75 
= NLeE Via 6. 3 v5, m+Vm2+1 
4 2 10. 
3m2n 
Page 216. 
5+ ivi. g Stiv2, 5, Stiv7l, 
2 3 6 
—9+iv7 g Ut iv103, 6. 
4 14 ; 2 
"7. 8 4 £1.08, 8. o + tv 23 iV23. 
Ae) 
Page 217. First Exercise, no answer. 
Second Exercise. 
—~5+4+vV6i 7. +10. 154+ V117 
pata: “alin ieh 19) eee 
6 8. 0, — j. 4 
34+ V6. 25 + 5iv47. 181s 
3 18 14. 1, —1. 
24+V7. 10, ~2ZLt ¥501, 15, Lt 2iv2, 
i, — 3. Gir + 80 eee 
1, —4§, wu, 2£tv17, 16, Do£9V33, 
ey AG, 7 2 | 
Page 223. 


w—{e—-2=0, 2+2%—20=0, 22?7—1824+1=0, 

6. zc? -—62+11=0. 

7. 14422 — 864”%+4 1171 =0. 
8. x7-—2ax+a?+b?=0. 
9. 27-—61lx+ 912+ 25m =0. 
10. a? —2V5ax + 48 = 0, 


7 — 245 —4= 0. 
v*—6x+34=—0. 


. de?7—-4xe—-55=0. 








25 


ANSWERS. 
Page 225. 
1 ~8tVB -38tV13_ 11. 4+ 5, 4 4% 
2 4 2 19.80%: 
Beate Wd N/S 13. 18. 
[pie 2 cas ais 14.14V4 ¥ 2Vv3i. 
ota — 6 + 2V'89 | Tyee: —145Vv5_ 
iy 15 2 
5. =Aa tive, 16. 4, —9, — SE VOL, 
bi 2S, Lats, Vio 410, =a 
7 — 33, —3. 18. 9. 
8. +37, + V2. 19. 16, 5V5. 
9. +2V2i, +i. , 384+V505 
3/= M4 20. 2, Oe) 
10. 8vV3; 227-4. 4 
Page 226. 
Peele teil 2k., 14. 4 mi. 
SS. 15. 24 min., 30 min. 
3. 18, 12. 16. A’s, 5 mi.; B’s, 3 mi. 
4. 30; 11. 172 10;A; 
a . mh i na We = 04 ve Wee 
: ’ 9 495 to, — 14, — 11. 
7. 100 ft. by 132 ft. Le Vins eee 
8. 20. 20. 6 mi., 3 mi. 
9. 15. 21. 15 ft., 20 ft., 25 ft. 
1Oredy 20) 3-20, — 12. 99, et Va'—4ab a~Va—4 ab 
11. 60 sq. ft. 2 ; 2 
12. 39,40; — 40, — 39. 23. 24 rd. by 60 rd. 
Pit a) 23. — 21; 24. 10 hr. 
Page 235. 
1, *(0, 5),-(6, 0). Sin GL 8) Cae) 
2. (0, 2), (3, 0). On" 452), Co hi ©) 
Patan). (—6, — 8). LONE 72 10): 
4. (7, 8), (8, 7). uw. (ee 4v3E ets! : 
Barto ),4--15, — 9). AIS Oca i. Vaid in 
fee my, C6, == 9). P25 (105 )5 C1 Ea): 
7. (2,3), C1, 6). 


pb oe 


a. 
2. 


THE ESSENTIALS OF ALGEBRA. 


13, (6, 8), (4, 0). 

ra! (-} 1 2evR), (-} 1 —). 
3 6 3 6 

15. (— 3, 5), (—7, 7). 


Page 238. 


(2, 3), 2; —3), (— 2, 3), (= —2, —35). 

(3, 4), (5, —4), Co 3,4) a, a). eae ~ ou 
(4V5i, V105), (4V5i, —V105), (—4V57, V105), 
(—4V51, — V105). 

(3, 7), (3, Ei), (—3, 7), (Oy ale 

(-V203 i, V85), (V203 i, —V85), (—V 2087, V85), 
(— V203 i, — V85). 

(6,2); (6, 9), (5) 9) (bao 

(6V 3h 2V$r), (6V 3 -2V ’ (-6V3;, 2V35), 
C= 6V sr, —2V3,). 

(6V43, 6V 2,1), (6V$8, —6V 2,1), (6V43, 6VE, HO; 
(—6V$3, —6V 7%, 1). 











Page 241. 


(2,1), (—2, —1), (1, 2), (-1, —2). 

(Vii, Sawa 4) CWA, 3Vi i), (SV, 5Vae)s 
ez —8V4h, —5V 4). 

(3, 4), (—8, —4), (7V$, 2V4), (-7V§, -—2V9). 
2, 1), (—2, —1)s (4V4, v4); (—4v4, — v4). 
(8V3, v3), (—3 v3, —v3), (4, 5), (—4, —~ 0) 
(5, —L); (-5, 1), (3, =); (—3, 4). 

(2,4), (2; —4), (V2, 3V2), (1 ee 
(4°) (46): 


Cd, 5), Ce 1, =); (4V42i, V#2 4), (—4vV42i, —v}2%). 


lem — bn, [an — Gf 
Ay ———— ————. 
* Fee oe am — Ib 


: G; 3) (—-$, —- » (4 3), Gey 3 ’ — $). 


(3, 5), (—3, 165; (8, 5), Ce 8, —5). 


Page 245. 


4. (28))-( 2) — 2h eye Cee 
2. (8, 5), (8; — By (5) 3)) (oem 


OTSws ooo ie 


ANSWERS. QT 





Br (o8)..(8, 3). 8. (8, 5), (4 10). 
4. (3,3), (3, 8). 9. Gb, Gd- 
5. (7, 2), (—2, —7). 10. (4, 4). 
6 (4, 5), (5):8). Tete 8) oe). 
Pans, —3). 1246, 3):2(57 3): 

13. (10, 20), (20, 10). 

a+b a—b a—b a+b 
eae) a) 
Page 249, 

(4, 3), (—4%, a) 4. (2, 1), (1, 2). 7. (3, 1), (1, 3). 
(4, 2), (5, 4). 5. (3, 2), (—2, —3). 8. (3, 2), (2, 3). 
(1, 2), (14, 4). 6. (4, 3), (3, 4). 9. (38, 2), (—4, —4). 


fee ey 1). (8,4), (4, 3); (—6, —2), (—2; —6). 
PeewotehVe1t\ | flaVoleealev sii 

(3, 2), (—2; —3), (ee ey ey Se ok 

fetta), (24 57,2 — 51), (2— 57, 24.5%). 

(2, 5), (—2, —5), (8V3, 7V4), (—8 V4, —TV}). 

(2, 1), (— 2, —1), es 4), — 3, ey t). 


(3, 9), (9, 8). ‘VW. Cio ZY (-4.\ 20-0); 
(1, 6), (6, 1), (Met BN, (25M Sev, 
2 2 2 
(4, 2), (—2, — 4). 20. (2, 3), (, 2). 
Vie: 6. 5; 2. 11. 66 ft. by 200 ft. 
25, 21. 7. 900sq.rd.,1600sq.rd. 12. 48 or 34, 
3,4 8. 50 rd. by 20 rd. 13. 4 yd., 5 yd. 
6, 11. 9. 50 ft. by 120 ft. 14. 20 da., $3. 
5, 3 10. 8, 16. 15.44, 
Page 251. 
cy —iay —2a?. 
- (10, By (— 12, 5), (4a, 5a), (8, —34), (-—c+a,c¢—Db). 
“2 —62x+4 2. 10. (— rary 
137 V3 — 56. 11 1+v1—4ced 
a8 + o2Va-1 — 1 — Va-8, ; od 
10 +. 8V30 — 3V6 —2V5. 13. (35.and 86) (— 86. and — 85). 
34 14. 85. 
5 + 127 15. 6,,. 
26 , 1G WiK6 A). Ge Gin 4). 
ad + add} + 03. 17. (3, 2), (— 3, 


re 


ay Pd heer 











THE ESSENTIALS OF ALGEBRA. 


52 mi., 40 mi. oa ere 5a 
OF Do 4’ G+ 

=the VSB 25. Ont 6 ak URateee 
Hy ae 26. 16. 
ion gE a 27. bd. 

Vy ys 28. 2,1, 2, 10. 

+iv3, +4. 29. 450 abay. ei 

30. 8+y-H = 2V8¢g. 





V(11)?, 4. us 
é: ) 31. 4 b. 
. 26, 52. a+b 
4 3 2.3 2 5 8 5 
82. a + b2 +064 2a8bt — 2a8c? —2dtc?. 
11+ 4V6. 40. 2, 5. 
35 + V1049 41, CHP 4V2 a +204 
Land , a? — b» 
@2@ — 15)(2a +38) 42. 144. py 
Ta(a — 3) 43 —15+V161. 
140 ft. by 200 ft. 8 
—2y. 44. 0. 
+S, + 3, 45. 40rd. by 40rd. 20rd. by 80rd. 
(1; 3)6 =) 8): | 
Page 259. 
$, 6: a—b ll. 2. 16. 4+ 8. 
4 4 ab hb Awe by ei 
¥ 7 a-—2 13 hou Lesa: 
a2, 8. 224+ 5a. 2 7 19:1 om 
x—-Y-—2. 9 at+ay+y*% 14 1. 20. 7. 
x—Y. LOWE. 15. x«—1. 
Page 262. 
60. 3. 33. 4. 75.3984. 9. + V10. 10. 33511.1. 
Page 266. 
113 5. + aN ene 1 8. 49, 22, 14. ac 
37° 6. 53! b 9.88; b 
+ 15, anit ae 10. Yes. 15, i, 
— $$. EE Nha 11. No. ‘True b 
forz=1. 
Page 273. Section 194. 
120. 3. 1260. 5. 12,870. 7. 13992. 9. 91,390. 
6. 4. 58,140. 6. 2. 8. 300. 10. +45. 


ANSWERS. Se RD AL, 


Section 195. 


17.120. 5. 75,600. 9. 120. 
2. 840, 5040. 6. 720. 10. 24. 

3. 3360; 57,120; 1,860,480. Mae BI GON 1 ZU: il. 80,640. 
4. 358,800. 8. 630; 34,650. 12. 27,720. 
Page 2777. 

184-126. 3. 15,504. 5. 4275. rf Pyee ba 9. 840. 


2. 210; 220; 220. 4. 475. 6. 1001. 8. 336; 456. 10. 16. 


1. 53. 4, 365. 
2. 144. 5. 20. 
3. 14, 18, 22, 26, 30. 


\6. 6, 11, 16, 21, etc. 


Reo ka 
2. 1023. 


8. 6561, 9840. 6. 


= 


Page 283. 
6. 13, 483. 10. 30. 13. 7650. 
8. —75. A Eo Oly 14. 2745. 
9. 135. 12. 78. 15. $405. 


772 50) 268, 482d) ot) 18015, 146: 


Page 288. 
4043, 7%. $20,475. 10. 8, 18. 
— 4920. 8. 5. 14. 8, 24, 72, 216. 
$255. 9. 14. 15. 2, :4, 8. 

Page 289. 


(1+18a% + 153 #2 + 816 23 + 3060 2 + 8568 a5+ 
2. | 1—21y + 210 y? — 1330 y? + 5985 yt — 20349 y+ 
(25 4+ 25 x24y + 300 x23y? 4+ 2300 x22y3 + 12650 a21y4 + 53130 129yo+ 


1x2x3x.-.-15 


30 x 29 x ---16 ‘ 


((a) 1+%+27+ 23+ xt+ 

(0) 1444-42? + apa — hy ott 
| (cy) Lt gat yiat + Hh? + pag att 

(a) 1— at §a2— 490 + Hg ot 


— 


Page 290. 


64 28 — 192 xy + 240 aty2 — 160 x3y3 + 60 x2yt — 12 xy® + yw, 


x + 10 vty + 40 wy? + 80 xy? + 80 ryt + 32 ¥?. 
* | 128 a? — 1344 abd + 6048 a5b? — 15120 a'b? + 22680 abt 


[ §  _ 90412 a2b® + 10206 ab® + 2187 b7. 


co 


THE ESSENTIALS OF ALGEBRA. 





256 x8 — 1 6. 2, 6, 18 11. 293,930. 
2”%—1 7 +16. 12. 59,400. 
1950. 8. 27. 13. 5040. 
672 x. 9. — 252 x 32 x 243. 14. 1980. 
saa 10. £433. 
18. 2h ere te — rot 





= Se 2 oe 2) eee 
16. 10 20 E000 — 1800000 1280000000° 


17. 1+6y%+15%+20% v2 +15 a 46 ania 
$ 


20. 
25. 


rm 


i 
= 


Ene er 


ge ep LT le lee od a 


b2 b2 Vb B3 
18,564. 21. $ 465 ; $3075. 
A TEEY 22. 23. 
Lote 6. 24. 8a*—S8ava?—1—4. 
1+65%+ 152724 30234 4524 + 51 2 4+ 45 xo “ 30 x7 + 1528 
+ 5x9 + x, 
Page 296. 
2. Sopa 5. 0. U8! 9. 5 
1. Alene 6.) = 3. 8. 2. 10. 3 
Page 297. 
5.6590. 6. 4.8663. 11. 1.3010; 1.8751. 
3.9943. 7. .9030; 1.2040; 1.5050. 12. 1.6902; 2.5353. 
a At 7627 8. .9542 ; 1.4313. 18. 1.6232. 
0.5065. 9. 1781 ; 1.3801. 14. 2.1461. 
6.8482. 10. 1.3980; 2.0970. 15. 1.6620. 
Page 302. 
2.7292. 11. 0.4971. 21. 651. 31. .592. 
2.9930. 12. 0.43848. 22. 82.13. 32. .0144. 
2.7042. 13. 0.1504. OS hay: 33. .00391. 
3.1287. 14. 1.0969. 24. 5890. 34. .0446. 
1.3075. 15. 1.5740. 25. 14.8. 35. .293. 
2.07 Tl: 16. 1.5740. 26. 2.80. 36. .30087. 
0.1976. 17. 2.9931. 27. 143.74. 37. .33392. 
2.8319. 18. 0.7321 28. 1812.9. 38. .00248, 
4.7482. 19. 1.5981. 29. 2.3747. 39. .30871. 
1.7190. 20. 3.7691. 30. 39.509. 40. .039682. 


oS 


_= 


719500. 


762400000. 


254. 


4.39. 
2.52. 


$ 2429.40. 
$ 3165.70. 


ANSWERS. 
Page 305. 
4. 19.014. 7. 9.556. 
5. 61900. 8. 1138.05. 
6G. 1004.6. 
Page 306. 
3.7 622.23 Tae 2. 
1.59 6. 4.54 8. 3.59. 
Page 308. 
3. 12.86 yr. 5. 23.51 yr. 
4. $1258.60. 6. $50.12. 


31 


9. 8.9975. 
10. 1221100. 


7. 3.55%. 
8. $668.17. 











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THE ESSENTIALS OF ALGEBRA NEW YORK 


512.9AL2E 





